Question : If $\tan\theta =\frac{3}{4}$, then the value of $\frac{4\sin^{2}\theta–2\cos^{2}\theta}{4\sin^{2}\theta+3\cos^{2}\theta}$ is equal to:
Option 1: $\frac{1}{21}$
Option 2: $\frac{2}{21}$
Option 3: $\frac{4}{21}$
Option 4: $\frac{8}{21}$
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Correct Answer: $\frac{1}{21}$
Solution : $\frac{4\sin^{2}\theta–2\cos^{2}\theta}{4\sin^{2}\theta+3\cos^{2}\theta}$ Dividing this numerator and denominator by $\cos\theta$ we have, $\frac{4\sin^{2}\theta–2\cos^{2}\theta}{4\sin^{2}\theta+3\cos^{2}\theta} = \frac{4\tan^{2}\theta–2}{4\tan^{2}\theta+3}$ Putting the value of $\tan\theta =\frac{3}{4}$ in this expression, we have, = $\frac{4×\frac{9}{16}–2}{4×\frac{9}{16}+3}$ = $\frac{\frac{9}{4}–2}{\frac{9}{4}+3}$ = $\frac{\frac{1}{4}}{\frac{21}{4}}$ = $\frac{1}{21}$ Hence, the correct answer is $\frac{1}{21}$.
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Question : The value of $\frac{\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}$ is equal to:
Option 1: $\sin\theta$
Option 2: $\cos\theta$
Option 3: $\tan\theta$
Option 4: $\cot\theta$
Question : If $5\tan\theta=4$, then $\frac{5\sin\theta-3\cos\theta}{5\sin\theta+2\cos\theta}$ is equal to:
Option 1: $\frac{2}{3}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{6}$
Option 4: $\frac{1}{3}$
Question : If $\sin \theta-\cos \theta=0$, then what is the value of $\sin ^2 \theta+\tan ^2 \theta$ ?
Option 1: $\frac{1}{2}$
Option 2: $1$
Option 3: $\frac{4}{5}$
Option 4: $\frac{3}{2}$
Question : What is the value of $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}+\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}$?
Option 1: $\frac{1}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 2: $2\left(\sin ^2 \theta-\cos ^2 \theta\right)$
Option 3: $\frac{2}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 4: $\sin ^2 \theta-\cos ^2 \theta$
Question : If $3 \tan \theta=2$, what is the value of $\frac{(3 \sin \theta-\cos \theta)}{(3 \sin \theta+\cos \theta)}?$
Option 1: $1$
Option 2: $3$
Option 3: $\sqrt3$
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