Match List – I with List – II.
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LIST I |
LIST II |
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A. If $4 \sin ^{-1} x+\cos ^{-1} x=\pi$, then x equal to |
I. $\frac{\pi}{2}$ |
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B. The value of $\frac{1-\tan ^2 15°}{1+\tan ^2 15°}$ is |
II. $\frac{1}{2}$ |
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C. If $x+\frac{1}{x}=2$, then principal value of $\sin ^{-1} x$ is |
III. $\frac{3 \pi}{4}$ |
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D. Two angles of a triangle are $\cot ^{-1} 2$ and $\cot ^{-1} 3$, then third angle is |
IV. $\frac{\sqrt{3}}{2}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-IV, C-I, D-III
(A) $4 \sin ^{-1} x+\cos ^{-1} x=\pi$
$⇒4\sin ^{-1} x+\left(\frac{π}{2}-\sin ^{-1} x\right)=\pi$
$⇒3\sin ^{-1} x=\frac{π}{2}$
$⇒\sin ^{-1} x=\frac{π}{6}$
$⇒x=\sin\left(\frac{π}{6}\right)=\frac{1}{2}$ (II)
(B) $\frac{1-\tan ^2 15°}{1+\tan ^2 15°}=\tan(45°-15°)=\tan 30°$
$⇒\tan 30°=\frac{\sqrt{3}}{2}$ (IV)
(C) $x+\frac{1}{x}=2$
$⇒x^2-2x+1=0$
$⇒(x-1)^2=0$
$⇒x=1$
$∴\sin^{-1}(1)=\frac{\pi}{2}$ (I)
(D) $θ_3=\pi-(θ_1+θ_2)$
$=\pi-(\cot^{-1}2+\cot^{-1}3)$
$=\pi-\left(\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}\right)=\pi=\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}×\frac{1}{3}}\right)$
$=\pi-\tan^{-1}1=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$ (III)