Exercise 3.4 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 3.4
Question 1.
If $\sin x=\frac{15}{17}$ and $\cos y=\frac{12}{13}, 0 (ii) $\cos (x-y)$
(iii) $\tan (x+y)$.
Solution:
$\sin x=\frac{15}{17} ; x$ is in I quadrant $\cos y=\frac{12}{13}$

$y$ is in I quadrant
From $\quad \triangle \mathrm{ABC}, \mathrm{BC}=\sqrt{17^2-15^2}=\sqrt{289-225}$
$
=\sqrt{64}=8
$
From $\triangle \mathrm{PQR}, \mathrm{PQ}=\sqrt{13^2-12^2}=\sqrt{169-144}$
$
=\sqrt{25}=5
$

Now, $\quad \sin x=\frac{15}{17}, \quad \cos x=\frac{8}{17}: \tan x=\frac{15}{8}$ $\sin y=\frac{5}{13}, \quad \cos y=\frac{12}{13}: \tan y=\frac{5}{12}$
$
\text { (i) } \begin{aligned}
\sin (x+y) & =\sin x \cos y+\cos x \sin y \\
& =\frac{15}{17} \times \frac{12}{13}+\frac{8}{17} \times \frac{5}{13}
\end{aligned}
$

$
=\frac{180}{221}+\frac{40}{221}=\frac{220}{221}
$
(ii) $\cos (x-y)=\cos x \cos y+\sin x \sin y$
$
\begin{aligned}
& =\frac{8}{17} \times \frac{12}{13}+\frac{15}{17} \times \frac{5}{13} \\
& =\frac{96}{221}+\frac{75}{221}=\frac{171}{221}
\end{aligned}
$
(iii) $\tan (x$
$
\begin{aligned}
& =\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8} \times \frac{5}{12}} \\
& =\frac{45+10}{24} / 1-\frac{75}{96}=\frac{55}{24} / \frac{96-75}{96} \\
& =\frac{55}{24} / \frac{21}{96}=\frac{55}{24} \times \frac{96}{21}=\frac{220}{21}
\end{aligned}
$
Question 2.
If $\sin A=\frac{3}{5}$ and $\cos B=\frac{9}{41}, 0 (ii) $\cos (\mathbf{A}-\mathrm{B})$.
Solution:

$
\begin{aligned}
& \sin \mathrm{A}=\frac{3}{5} \\
& 0<\mathrm{A}<\pi / 2 \\
& \text { From } \triangle \mathrm{ABC}, \mathrm{AB}=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4 \\
& \cos \mathrm{B}=\frac{9}{41} \\
& 0<\mathrm{B}<\pi / 2
\end{aligned}
$

From $\triangle \mathrm{BAD}$
$
\begin{aligned}
\mathrm{AD} & =\sqrt{41^2-9^2}=\sqrt{(41+9)(41-9)} \\
& =\sqrt{50 \times 32}=\sqrt{100 \times 16} \\
& =\sqrt{10^2 \times 4^2}=10 \times 4=40
\end{aligned}
$

Now,
From $\triangle A B C, \sin \mathrm{A}=\frac{3}{5} ; \quad \cos \mathrm{A}=\frac{4}{5}$
From $\triangle A B D, \sin B=\frac{40}{41} ; \quad \cos B=\frac{9}{41}$

$(i)$
$
\begin{aligned}
\sin (A+B) & =\sin A \cos B+\cos A \sin B \\
& =\left(\frac{3}{5} \times \frac{9}{41}\right)+\left(\frac{4}{5} \times \frac{40}{41}\right) \\
& =\frac{27}{205}+\frac{160}{205}=\frac{187}{205}
\end{aligned}
$
(ii)
$
\begin{aligned}
\cos (A-B) & =\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B} \\
& =\left(\frac{4}{5} \times \frac{9}{41}\right)+\left(\frac{3}{5} \times \frac{40}{41}\right) \\
& =\frac{36}{205}+\frac{120}{205}=\frac{156}{205}
\end{aligned}
$
Question 3.
Find $\cos (x-y)$, given that $\cos x=-\frac{4}{5}$ with $\pi $
\pi $

Solution:
$
\begin{aligned}
& \cos x=-4 / 5 \\
& \pi \end{aligned}
$
$\Rightarrow x$ is in III quadrant
From $\triangle \mathrm{PQR}, \mathrm{PQ}=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3$

since $x$ is in III quadrant
Both $\sin x$ and $\cos x$ are negative
$\therefore \sin x=-\frac{3}{5}$ and $\cos x=-\frac{4}{5}$
$\sin y=-\frac{24}{25}$ and $y$ is in III quadrant
Both $\sin y$ and $\cos y$ are negative
$
\text { From } \begin{aligned}
\triangle \mathrm{ABC}, \mathrm{BC} & =\sqrt{25^2-24^2}=\sqrt{625-576} \\
& =\sqrt{49}=7
\end{aligned}
$

So, $\sin y=-\frac{24}{25}$ and $\cos y=-\frac{7}{25}$
Now $\cos (x-y)=\cos x \cos y+\sin x \sin y$
$=\left(-\frac{4}{5}\right)\left(-\frac{7}{25}\right)+\left(-\frac{3}{5}\right)\left(-\frac{24}{25}\right)$
$=\frac{28}{125}+\frac{72}{125}=\frac{100}{125}=\frac{4}{5}$

Question 4.
Find $\sin (x-y)$, given that $\sin x=\frac{8}{17}$ with $0

Solution:

$
\begin{aligned}
& \sin x=\frac{8}{17}, 0 & \therefore \sin x, \cos x \text { are }+\mathrm{ve} \\
& \text { From } \triangle \mathrm{ABX}, \mathrm{AX}=\sqrt{17^2-8^2}=\sqrt{(17+8)(17-8)} \\
& =\sqrt{(25)(9)}=5 \times 3=15 \\
& \therefore \sin x=\frac{8}{17} \text { and } \cos x=\frac{15}{17} \\
& \cos y=-\frac{24}{25}, \pi & \text { So, } \sin y \text { and } \cos y \text { are -ve } \\
& \text { From } \triangle \mathrm{ALY}, \mathrm{AL}=\sqrt{25^2-24^2} \\
& =\sqrt{49}=7 \\
& \therefore \cos y=-\frac{24}{25} \text { and } \sin y=-\frac{7}{25} \\
& \sin (x-y)=\sin x \cos y-\cos x \sin y \\
& =\left(\frac{8}{17}\right)\left(-\frac{24}{25}\right)-\left(\frac{15}{17}\right)\left(-\frac{7}{25}\right) \\
& =-\frac{192}{425}+\frac{105}{425}=-\frac{87}{425} \\
&
\end{aligned}
$
Question 5.
Find the value of $(i) \cos 105^{\circ}$
(ii) $\sin 105^{\circ}$
(iii) $\tan \frac{7 \pi}{12}$
Solution:

$
\begin{aligned}
& 105^{\circ}=60^{\circ}+45^{\circ} \\
& \text { (i) } \cos 105^{\circ}=\cos \left(60^{\circ}+45^{\circ}\right) \\
& =\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ} \\
& =\frac{1}{2} \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}}=\frac{1}{2 \sqrt{2}}-\frac{\sqrt{3}}{2 \sqrt{2}} \\
& =\frac{1-\sqrt{3}}{2 \sqrt{2}} \\
&
\end{aligned}
$
$
\text { (ii) } \begin{aligned}
\sin 105^{\circ} & =\sin \left(60^{\circ}+45^{\circ}\right) \\
& =\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ} \\
& =\frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}}+\frac{1}{2} \frac{1}{\sqrt{2}} \\
& =\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}
\end{aligned}
$
(iii) Now $\tan \frac{7 \pi}{12}=\tan \frac{7}{12} \times 180^{\circ}=\tan 7 \times 15^{\circ}=\tan 105^{\circ}$
$
\begin{aligned}
& =\frac{1+\sqrt{3}}{2 \sqrt{2}} / \frac{1-\sqrt{3}}{2 \sqrt{2}} \\
& =\frac{1+\sqrt{3}}{2 \sqrt{2}} \times \frac{2 \sqrt{2}}{1-\sqrt{3}}=\frac{1+\sqrt{3}}{1-\sqrt{3}} \\
& =\frac{1+\sqrt{3}}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}} \\
& =\frac{1+\sqrt{3}+\sqrt{3}+3}{1-3}=\frac{4+2 \sqrt{3}}{-2} \\
& =\frac{2(2+\sqrt{3})}{-2}=-(2+\sqrt{3})
\end{aligned}
$
Question 6.
Prove that (i) $\cos \left(30^{\circ}+x\right)=\frac{\sqrt{3} \cos x-\sin x}{2}$ (ii) $\cos (\pi+\theta)=-\cos \theta$
(iii) $\sin (\pi+\theta)=-\sin \theta$.

Solution:
(i) $\mathrm{LHS}=\cos \left(30^{\circ}+x\right)=\cos 30^{\circ} \cos x-\sin 30^{\circ} \sin x$
$
\begin{aligned}
& =\frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x \\
& =\frac{\sqrt{3} \cos x-\sin x}{2}=\mathrm{RHS}
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) L.HS }=\cos (\pi+\theta)=\cos \left(180^{\circ}+\theta\right) \\
& =\cos 180^{\circ} \cos \theta-\sin 180^{\circ} \sin \theta \\
& =(-1) \cos \theta-(0) \sin \theta \\
& =-\cos \theta=\text { RHS }
\end{aligned}
$
(iii) LHS $=\sin (\pi+\theta)=\sin \pi \cos \theta+\cos \pi \sin \theta$
$=(0) \cos \theta+(-1) \sin \theta$
$
=-\sin \theta=\text { RHS }
$
Question 7.
Find a quadratic equation whose roots are $\sin 15^{\circ}$ and $\cos 15^{\circ}$

Solution:

$
\begin{aligned}
\sin 15^{\circ} & =\sin \left(45^{\circ}-30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\
& =\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \frac{1}{2}=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}} \\
& =\frac{\sqrt{3}-1}{2 \sqrt{2}}=\frac{\sqrt{3}-1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4} \\
\cos 15^{\circ} & =\cos \left(45^{\circ}-30^{\circ}\right)=\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ} \\
& =\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \frac{1}{2}=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}} \\
& =\frac{\sqrt{3}+1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}
\end{aligned}
$
Now the quadratic equation with roots $\cos 15^{\circ}$ and $\sin 15^{\circ}$ is
$
x^2-\left(\cos 15^{\circ}+\sin 15^{\circ}\right) x+\left(\cos 15^{\circ} \sin 15^{\circ}\right)=0
$
Now, $\quad \cos 15^{\circ}+\sin 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}$
$
\begin{aligned}
& =\frac{2 \sqrt{6}}{4}=\frac{\sqrt{6}}{2} \\
\cos 15^{\circ} \times \sin 15^{\circ} & =\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right) \\
& =\frac{6-2}{16}=\frac{4}{16}=\frac{1}{4}
\end{aligned}
$
Now the required quadratic equations
(i.e) $4 x^2-2 \sqrt{6} x+1=0$
$
x^2-\left(\frac{\sqrt{6}}{2}\right) x+\left(\frac{1}{4}\right)=0
$

Question 8.
Expand $\cos (\mathrm{A}+\mathrm{B}+\mathrm{C})$. Hence prove that $\cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}=\sin \mathrm{A} \sin \mathrm{B} \cos \mathrm{C}+\sin \mathrm{B} \sin \mathrm{C} \cos \mathrm{A}+$ $\sin \mathrm{C} \sin \mathrm{A} \cos \mathrm{B}$, if $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2}$
Solution:
Taking $\mathrm{A}+\mathrm{B}=\mathrm{X}$ and $\mathrm{C}=\mathrm{Y}$
We get $\cos (\mathrm{X}+\mathrm{Y})=\cos \mathrm{X} \cos \mathrm{Y}-\sin \mathrm{X} \sin \mathrm{Y}$
(i.e) $\cos (A+B+C)=\cos (A+B) \cos C-\sin (A+B) \sin C$
$=(\cos A \cos B-\sin A \sin B) \cos C-[\sin A \cos B+\cos A \sin B] \sin C$
$\cos (A+B+C)=\cos A \cos B \cos C-\sin A \sin B \cos C-\sin A \cos B \sin C-\cos A \sin B \sin C$ If (A+
$B+C)=\pi / 2$ then $\cos (A+B+C)=0$
$
\begin{aligned}
& \Rightarrow \cos A \cos B \cos C-\sin A \sin B \cos C-\sin A \cos B \sin C-\cos A \sin B \sin C=0 \\
& \Rightarrow \cos A \cos B \cos C=\sin A \sin B \cos C+\sin B \sin C \cos A+\sin C \sin A \cos B
\end{aligned}
$

Question 9
. Prove that
(i) $\sin \left(45^{\circ}+\theta\right)-\sin \left(45^{\circ}-\theta\right)=\sqrt{2} \sin \theta$.
(ii) $\sin \left(30^{\circ}+\theta\right)+\cos \left(60^{\circ}+\theta\right)=\cos \theta$.
Solution:
(i)
$
\begin{aligned}
& \sin \left(45^{\circ}+\theta\right)=\sin 45^{\circ} \cos \theta+\cos 45^{\circ} \sin \theta \\
& =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\
& \sin \left(45^{\circ}-\theta\right)=\sin 45^{\circ} \cos \theta-\cos 45^{\circ} \sin \theta \\
& =\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta \\
& \text { From (1), (2) } \Rightarrow \text { LHS }=\sin \left(45^{\circ}+\theta\right)-\sin \left(45^{\circ}-\theta\right) \\
& =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\
& =\frac{2}{\sqrt{2}} \sin \theta=\frac{2 \sqrt{2}}{\sqrt{2} \sqrt{2}} \sin \theta=\sqrt{2} \sin \theta=\text { RHS } \\
&
\end{aligned}
$

(ii)

$\begin{aligned}
\sin \left(30^{\circ}+\theta\right) & =\sin 30^{\circ} \cos \theta+\cos 30^{\circ} \sin \theta \\
& =\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta \\
\cos \left(60^{\circ}+\theta\right) & =\cos 60^{\circ} \cos \theta-\sin 60^{\circ} \sin \theta \\
& =\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta \\
(3)+(4) \Rightarrow \text { LHS }=\sin \left(30^{\circ}\right. & +\theta)+\cos \left(60^{\circ}+\theta\right)=\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta+\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta \\
& =\frac{1}{2} \cos \theta+\frac{1}{2} \cos \theta=\cos \theta=\text { RHS }
\end{aligned}$
Question 10.
If $a \cos (x+y)=b \cos (x-y)$, show that $(a+b) \tan x=(a-b) \cot y$.
Solution:
$
\begin{aligned}
& a \cos (x+y)=b \cos (x-y) \\
& a[\cos x \cos y-\sin x \sin y]=6[\cos x \cos y+\sin x \sin y] \\
& \text { (i.e) } a \cos x \cos y-a \sin x \sin y=b \cos x \cos y+b \sin x \sin y \\
& a \cos x \cos y-b \sin x \sin y=a \sin x \sin y+b \cos x \cos y
\end{aligned}
$

(i.e) $\frac{a \frac{\cos y}{\sin y}-b \frac{\sin x}{\cos x}}{\sin y \cos x}=\frac{a \frac{\sin x}{\cos x}+b \frac{\cos y}{\sin y}}{\sin y \cos x}$
$\Rightarrow a \cot y-b \tan x=a \tan x+b \cot y$
$a \cot y-b \cot y=a \tan x+b \tan x$
$\Rightarrow(a+b) \tan x=(a-b) \cot y$.
Question 11.
Prove that $\sin 105^{\circ}+\cos 105^{\circ}=\cos 45^{\circ}$.
Solution:
$
\begin{aligned}
& \sin 105^{\circ}=\sin \left(60^{\circ}+45^{\circ}\right) \\
&=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \cos 45^{\circ} \\
&=\frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}}+\frac{1}{2} \frac{1}{\sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \\
& \cos 105^{\circ}=\cos \left(60^{\circ}+45^{\circ}\right)=\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ} \\
&=\frac{1}{2} \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}}=\frac{1}{2 \sqrt{2}}-\frac{\sqrt{3}}{2 \sqrt{2}}=\frac{1-\sqrt{3}}{2 \sqrt{2}}
\end{aligned}
$
$
\begin{aligned}
& \text { So, } \mathrm{LHS}=\sin 105^{\circ}+\cos 105^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}+\frac{1-\sqrt{3}}{2 \sqrt{2}} \\
&=\frac{\sqrt{3}+1+1-\sqrt{3}}{2 \sqrt{2}}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}} \\
&=\cos 45^{\circ}=\text { RHS }
\end{aligned}
$
Question 12.
Prove that $\sin 75^{\circ}-\sin 15^{\circ}=\cos 105^{\circ}+\cos 15^{\circ}$.
Solution:
$
\begin{aligned}
& \text { RHS }=\cos 105^{\circ}+\cos 15^{\circ}=\cos \left(90^{\circ}+15^{\circ}\right)+\cos \left(90^{\circ}-75^{\circ}\right) \\
& =-\sin 15^{\circ}+\sin 75^{\circ} \\
& =\sin 75^{\circ}-\sin 15^{\circ}=\text { LHS }
\end{aligned}
$
Question 13.
Show that $\tan 75^{\circ}+\cot 75^{\circ}=4$

Solution:
$
\begin{aligned}
\tan 75^{\circ} & =\tan \left(45^{\circ}+30^{\circ}\right)=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}} \\
& =\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}} / \frac{\sqrt{3}-1}{\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \\
\cot 75^{\circ} & =\frac{1}{\tan 75^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}
\end{aligned}
$
So, $\mathrm{LHS}=\tan 75^{\circ}+\cot 75^{\circ}$
$
\begin{aligned}
& =\frac{\sqrt{3}+1}{\sqrt{3}-1}+\frac{\sqrt{3}-1}{\sqrt{3}+1} \\
& =\frac{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{3+1+2 \sqrt{3}+3+1-2 \sqrt{3}}{\sqrt{3}^2-1^2} \\
& =\frac{8}{3-1}=\frac{8}{2}=4=\text { RHS }
\end{aligned}
$

Question 14.
Prove that $\cos (\mathrm{A}+\mathrm{B}) \cos \mathrm{C}-\cos (\mathrm{B}+\mathrm{C}) \cos \mathrm{A}=\sin \mathrm{B} \sin (\mathrm{C}-\mathrm{A})$.
Solution:
$
\begin{aligned}
& \text { LHS }=(\cos A \cos B-\sin A \sin B) \cos C-(\cos B \cos C-\sin B \sin C) \cos A \\
& =\cos A \cos B \cos C-\sin A \sin B \cos C-\cos A \cos B \cos C+\cos A \sin B \sin C \\
& =\cos A \sin B \sin C-\sin A \sin B \cos C \\
& =\sin B[\sin C \cos A-\cos C \sin A] \\
& =\sin B[\sin (C-A)]=\text { RHS }
\end{aligned}
$
Question 15.
Prove that $\sin (\mathrm{n}+1) \theta \sin (\mathrm{n}-1) \theta+\cos (\mathrm{n}+1) \theta \cos (\mathrm{n}-1) \theta=\cos 2 \theta, \mathrm{n} \in \mathrm{Z}$.
Solution:
Taking $(\mathrm{n}+1) \theta=\mathrm{A}$ and $(\mathrm{n}-1) \theta=\mathrm{B}$
$\mathrm{LHS}=\sin \mathrm{A} \sin \mathrm{B}+\cos \mathrm{A} \cos \mathrm{B}$
$=\cos (\mathrm{A}-\mathrm{B})$
$=\cos [(\mathrm{n}+1)-(n-1)] \theta$
$=\cos (\mathrm{n}+1-\mathrm{n}+1) \theta=\cos 2 \theta=\mathrm{RHS}$

.Question 16
If $x \cos \theta=y \cos \left(\theta+\frac{2 \pi}{3}\right)=z \cos \left(\theta+\frac{4 \pi}{3}\right)$, find the value of $x y+y z+z x$.
Solution:
$
\begin{aligned}
& x \cos \theta=y \cos \left(\theta+\frac{2 \pi}{3}\right)=z \cos \left(\theta+\frac{4 \pi}{3}\right)=k \text { (say) } \\
& \frac{k}{x}=\cos \theta \\
& \frac{k}{y}=\cos \left(\theta+\frac{2 \pi}{3}\right) \\
& \frac{k}{z}=\cos \left(\theta+\frac{4 \pi}{3}\right) \\
& \frac{k}{x}+\frac{k}{y}+\frac{k}{z}==\cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{4 \pi}{3}\right) \\
& \frac{k}{x}+\frac{k}{y}+\frac{k}{z}=0 \\
& k\left[\frac{y z+x z+x y}{x y z}\right]=0 \Rightarrow x y+y z+z x=0
\end{aligned}
$

Question 17.
Prove that
(i) $\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})=\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}$
(ii) $\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})=\cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\cos ^2 \mathrm{~B}-\sin ^2 \mathrm{~A}$
(iii) $\sin ^2(A+B)-\sin ^2(A-B)=\sin 2 A \sin 2 B$
(iv) $\cos 8 \theta \cos 2 \theta=\cos ^2 5 \theta-\sin ^2 3 \theta$
Solution:
$
\begin{aligned}
& \text { (i) } \mathrm{LHS}=\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B}) \\
& =(\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B})(\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}) \\
& =\sin ^2 \mathrm{~A} \cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A} \sin ^2 \mathrm{~B} \\
& =\sin ^2 \mathrm{~A}\left(1-\sin ^2 \mathrm{~B}\right)-\left(1-\sin ^2 \mathrm{~A}\right) \sin ^2 \mathrm{~B} \\
& =\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~A} \sin ^2 \mathrm{~B}-\sin ^2 \mathrm{~B}+\sin ^2 \mathrm{~A} \sin ^2 \mathrm{~B} \\
& =\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\text { RHS }
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) LHS }=\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})=(\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B})(\cos \mathrm{A} \cos \mathrm{B}+\sin (\mathrm{A} \sin \mathrm{B}) \\
& =\cos ^2 \mathrm{~A} \cos ^2 \mathrm{~B}-\sin ^2 \mathrm{~A} \sin ^2 \mathrm{~B} \\
& =\cos ^2 \mathrm{~A}\left(1-\sin ^2 \mathrm{~B}\right)-\left(1-\cos ^2 \mathrm{~A}\right) \sin ^2 \mathrm{~B} \\
& =\cos ^2 \mathrm{~A}-\cos ^2 \mathrm{~A} \sin ^2 \mathrm{~B}-\sin ^2 \mathrm{~B}+\cos ^2 \mathrm{~A} \sin ^2 \mathrm{~B} \\
& =\cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\mathrm{RHS} \\
& \text { Now } \cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\left(1-\sin ^2 \mathrm{~A}\right)-\left(1-\cos ^2 \mathrm{~B}\right) \\
& =1-\sin ^2 \mathrm{~A}-1+\cos ^2 \mathrm{~B} \\
& =\cos ^2 \mathrm{~B}-\sin ^2 \mathrm{~A}
\end{aligned}
$

$\begin{aligned}
& \text { (iii) } \sin ^2 A-\sin ^2 B=\sin (A+B) \sin (A-B) \\
& L H S=\sin ^2(A+B)-\sin ^2(A-B)=\sin [(A+B)+(A-B)][\sin (A+B)-(A-B)] \\
& =\sin 2 A \sin 2 B=\text { RHS }
\end{aligned}$

$\begin{aligned}
& \text { (iv) LHS }=\cos 8 \theta \cos 2 \theta \\
& =\cos (5 \theta+3 \theta) \cos (5 \theta-3 \theta) . \\
& \text { We know } \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})=\cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~B} \\
& \therefore \cos (5 \theta+3 \theta) \cos (5 \theta-3 \theta)=\cos ^2 5 \theta-\sin ^2 3 \theta=\text { RHS }
\end{aligned}$

Question 19.
If $\cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\gamma-\alpha)=-\frac{3}{2}$, then prove that $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma$ Solution:
$
\begin{aligned}
& \Rightarrow \text { Given } \cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\gamma-\alpha)=\frac{-3}{2} \\
& 2 \cos (\alpha-\beta)+2 \cos (\beta-\gamma)+2 \cos (\gamma-\alpha)=-3 \\
& 2 \cos (\alpha-\beta)+2 \cos (\beta-\gamma)+2 \cos (\gamma-\alpha)+3=0 \\
& {[2 \cos \alpha \cos \beta+2 \sin \alpha \sin \beta]+[2 \cos \beta \cos \gamma+2 \sin \beta \sin \gamma]+[2 \cos \gamma \cos \alpha+\sin \gamma \sin \alpha]+3=0} \\
& =[2 \cos \alpha \cos \beta+2 \cos \beta \cos \gamma+2 \cos \gamma \cos \alpha]+[2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma+2 \sin \gamma \sin \alpha]+\left(\sin ^2 \alpha+\right. \\
& \left.\cos ^2 \alpha\right)+\left(\sin ^2 \beta+\cos ^2 \beta\right)+\left(\sin ^2 \gamma+\cos ^2 \gamma\right)=0 \\
& \Rightarrow\left(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+2 \cos \alpha \cos \beta+2 \cos \beta \cos \gamma+2 \cos \gamma \cos \alpha\right)+\left(\sin ^2 \alpha+\sin ^2 \beta\right)+\left(\sin ^2 \gamma+\right. \\
& 2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma+2 \sin \gamma \sin \alpha)=0 \\
& (\cos \alpha+\cos \beta+\cos \gamma)^2+(\sin \alpha+\sin \beta+\sin \gamma)^2=0 \\
& =(\cos \alpha+\cos \beta+\cos \gamma)=0 \operatorname{and} \sin \alpha+\sin \beta+\sin \gamma=0 \\
& \text { Hence proved }
\end{aligned}
$
Question 20.
Show that $(i) \tan \left(45^{\circ}+A\right)=\frac{1+\tan A}{1-\tan A}$
(ii) $\tan \left(45^{\circ}-A\right)=\frac{1-\tan A}{1+\tan A}$

Solution:
(i)
$
\begin{aligned}
\text { LHS }=\tan \left(45^{\circ}+\mathrm{A}\right) & =\frac{\tan 45^{\circ}+\tan \mathrm{A}}{1-\tan 45^{\circ} \tan \mathrm{A}} \\
& =\frac{1+\tan \mathrm{A}}{1-\tan \mathrm{A}}=\text { RHS }
\end{aligned}
$
(ii)
$
\begin{aligned}
\text { LHS }=\tan \left(45^{\circ}-\mathrm{A}\right) & =\frac{\tan 45^{\circ}-\tan \mathrm{A}}{1+\tan 45^{\circ} \tan \mathrm{A}} \\
& =\frac{1-\tan \mathrm{A}}{1+\tan \mathrm{A}}=\text { RHS }
\end{aligned}
$
Question 21.
Prove that $\cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}$.
Solution:
$
\begin{aligned}
\tan (A+B)= & \frac{\tan A+\tan B}{1-\tan A \tan B} \\
\text { LHS }=\cot (A+B)= & \frac{1}{\tan (A+B)}=\frac{1}{\frac{\tan A+\tan B}{1-\tan A \tan B}} \\
& =\frac{1-\tan A \tan B}{\tan A+\tan B}
\end{aligned}
$
$\div$ both $\mathrm{Nr}$ and $\mathrm{Dr}$ by $\tan \mathrm{A} \tan \mathrm{B}$
$
\begin{aligned}
& =\frac{1-\tan A \tan B}{\tan A \tan B} / \frac{\tan A+\tan B}{\tan A \tan B} \\
& =\frac{1}{\tan A \tan B}-1 / \frac{1}{\tan B}+\frac{1}{\tan A}=\frac{\cot A \cot B-1}{\cot B+\cot A}=\text { RHS }
\end{aligned}
$
Question 22.
If $\tan x=\frac{n}{n+1}$ and $\tan y=\frac{1}{2 n+1}$, find $\tan (x+y)$

Solution:
$
\begin{aligned}
\tan (x+y) & =\frac{\tan x+\tan y}{1-\tan x \tan y} \\
& =\frac{\frac{n}{n+1}+\frac{1}{2 n+1}}{1-\left(\frac{n}{n+1}\right)\left(\frac{1}{2 n+1}\right)} \\
& =\frac{n(2 n+1)+1(n+1)}{(n+1)(2 n+1)} / \frac{(n+1)(2 n+1)-n}{(n+1)(2 n+1)} \\
& =\frac{2 n^2+n+n+1}{2 n^2+n+2 n+1-n}=\frac{2 n^2+2 n+1}{2 n^2+2 n+1}=1
\end{aligned}
$
Question 23.
Prove that $\tan \left(\frac{\pi}{4}+\theta\right) \tan \left(\frac{3 \pi}{4}+\theta\right)=-1$
Solution:
$
\begin{aligned}
& \tan \left(\frac{\pi}{4}+\theta\right)=\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}=\frac{1+\tan \theta}{1-\tan \theta} \\
& \tan \left(\frac{3 \pi}{4}+\theta\right)=\frac{\tan \frac{3 \pi}{4}+\tan \theta}{1-\tan \frac{3 \pi}{4} \tan \theta}=\frac{(-1)+\tan \theta}{1-(-1) \tan \theta} \\
&=\frac{\tan \theta-1}{1+\tan \theta} \\
& \therefore \quad \tan \left(\frac{\pi}{4}+\theta\right) \tan \left(\frac{3 \pi}{4}+\theta\right)=\frac{1+\tan \theta}{1-\tan \theta} \times \frac{-1+\tan \theta}{1+\tan \theta} \\
&= \frac{\tan \theta-1}{1-\tan \theta}=-1=\text { RHS }
\end{aligned}
$

Question 24.
Find the values of $\tan (\alpha+\beta)$, given that $\cot \alpha=\frac{1}{2}, \alpha \in\left(\pi, \frac{3 \pi}{2}\right)$ and $\sec \beta=-\frac{5}{3}, \beta \in\left(\frac{\pi}{2}, \pi\right)$

Solution:

$
\begin{aligned}
& \cot \alpha=\frac{1}{2} \\
& \therefore \tan \alpha=\frac{1}{\cot \alpha}=2, \alpha \text { is in III quadrant } \\
& \sec \beta=-\frac{5}{3} \\
& \cos \beta=-\frac{3}{5}, \beta \text { is in III quadrant } \\
& \mathrm{AB}=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4 \\
& \therefore \tan \beta=\frac{\sin \beta}{\cos \beta}=\frac{4 / 5}{-3 / 5}=-\frac{4}{3} \\
& \therefore \beta \text { is in II quadrant, } \tan \beta=-\frac{4}{3} \\
& \text { Now } \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{2+\left(-\frac{4}{3}\right)}{1-(2)\left(-\frac{4}{3}\right)} \\
& =2-\frac{4}{3} / 1+\frac{8}{3}=\frac{6-4}{3} / \frac{3+8}{3} \\
& =\frac{2}{3} / \frac{11}{3}=\frac{2}{11} \\
&
\end{aligned}
$

Question 25.
If $\theta+\phi=\alpha$ and $\tan \theta=k \tan \phi$, then prove that $\sin (\theta-\phi)=\frac{k-1}{k+1} \sin \alpha$.
Solution:

$\begin{aligned}
& \theta+\phi=\alpha, \tan \theta=k \tan \phi \\
& k=\frac{\tan \theta}{\tan \phi} \\
& \frac{k-1}{k+1}=\frac{\frac{\tan \theta}{\tan \phi}-1}{\frac{\tan \theta}{\tan \phi}+1}=\frac{\tan \theta-\tan \phi}{\tan \theta+\tan \phi} \\
& =\frac{\frac{\sin \theta}{\cos \theta}-\frac{\sin \phi}{\cos \phi}}{\frac{\sin \theta}{\cos \theta}+\frac{\sin \phi}{\cos \phi}}=\frac{\sin \theta \cos \phi-\cos \theta \sin \phi}{\sin \theta \cos \phi+\cos \theta \sin \phi} \\
& \frac{k-1}{k+1}=\frac{\sin (\theta-\phi)}{\sin (\theta+\phi)}=\frac{\sin (\theta-\phi)}{\sin \alpha} \\
& \sin (\theta-\phi)=\frac{k-1}{k+1} \sin \alpha \\
&
\end{aligned}$