Question : Using trigonometric formulas, find the value of $(\frac{\sin (x-y)}{\sin (x+y)})(\frac{\tan x+\tan y}{\tan x-\tan y})$
Option 1: –2
Option 2: 2
Option 3: 0
Option 4: 1
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Correct Answer: 1
Solution : Given: $\frac{\sin (x-y)}{\sin (x+y)}(\frac{\tan x+\tan y}{\tan x–\tan y})$ Using $\sin(x+y)=\sin x\cos y+\cos x\sin y$ and $\sin(x-y)=\sin x\cos y-\cos x\sin y$, we get: $=(\frac{\sin x\cos y–\cos x\sin y}{\sin x\cos y+\cos x\sin y})(\frac{\tan x+\tan y}{\tan x–\tan y})$ Dividing both numerator and denominator by $\cos x\cos y$, we get, $=(\frac{\frac{\sin x\cos y–\cos x\sin y}{\cos x\cos y}}{\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}})(\frac{\tan x+\tan y}{\tan x–\tan y})$ $= (\frac{\tan x–\tan y}{\tan x+\tan y})(\frac{\tan x+\tan y}{\tan x–\tan y})$ $= 1$ Hence, the correct answer is 1.
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Question : If $\tan x = \frac{7}{5}$, the value of $\frac{9 \sin x – \frac{42}{5} \cos x}{15 \sin x + 21 \cos x}$ is:
Option 1: 0
Option 2: 1
Option 3: 0.1
Option 4: 0.5
Question : If $x\sin ^{2}60^{\circ}-\frac{3}{2}\sec 60^{\circ}\tan^{2}30^{\circ}+\frac{4}{5}\sin ^{2}45^{\circ}\tan ^{2}60^{\circ}=0$, then $x$ is:
Option 1: $-\frac{1}{15}$
Option 2: $–4$
Option 3: $-\frac{4}{15}$
Option 4: $–2$
Question : If $x=a(\sin\theta+\cos\theta), y=b(\sin\theta-\cos\theta)$, then the value of $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is:
Option 3: 2
Option 4: –2
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Option 1: 1
Option 2: 0
Option 4: 3
Question : If $x\sin^{3}\theta +y\cos^{3}\theta=\sin\theta\cos\theta$ and $x\sin\theta-y\cos\theta=0$, then the value of $\left ( x^{2}+y^{2} \right )$ equals:
Option 1: $1$
Option 2: $\frac{1}{2}$
Option 3: $\frac{3}2$
Option 4: $2$
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