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Exercise 4.6 · Q3

Q.sin⁡−135−cos⁡−11213+sec⁡−153−cosec−11312\sin^{-1}\dfrac35 - \cos^{-1}\dfrac{12}{13} + \sec^{-1}\dfrac53 - \text{cosec}^{-1}\dfrac{13}{12} is equal to

(1) 2π2\pi
(2) π\pi
(3) 00
(4) tan⁡−11265\tan^{-1}\dfrac{12}{65}
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✓ Free question

Using sec⁡−153=cos⁡−135\sec^{-1}\dfrac53=\cos^{-1}\dfrac35 and cosec−11312=sin⁡−11213\text{cosec}^{-1}\dfrac{13}{12}=\sin^{-1}\dfrac{12}{13}, the whole expression regroups into [sin⁡−135+cos⁡−135]−[cos⁡−11213+sin⁡−11213]\left[\sin^{-1}\tfrac35+\cos^{-1}\tfrac35\right]-\left[\cos^{-1}\tfrac{12}{13}+\sin^{-1}\tfrac{12}{13}\right], each bracket being exactly π2\dfrac{\pi}2.

Step 1. Convert the reciprocal-inverse terms. sec⁡−153=cos⁡−1(15/3)=cos⁡−135\sec^{-1}\dfrac53=\cos^{-1}\left(\dfrac1{5/3}\right)=\cos^{-1}\dfrac35, and cosec−11312=sin⁡−1(113/12)=sin⁡−11213\text{cosec}^{-1}\dfrac{13}{12}=\sin^{-1}\left(\dfrac1{13/12}\right)=\sin^{-1}\dfrac{12}{13}.

Step 2. Rewrite the full expression. sin⁡−135−cos⁡−11213+cos⁡−135−sin⁡−11213\sin^{-1}\dfrac35-\cos^{-1}\dfrac{12}{13}+\cos^{-1}\dfrac35-\sin^{-1}\dfrac{12}{13}.

Step 3. Regroup into complementary pairs. =[sin⁡−135+cos⁡−135]−[cos⁡−11213+sin⁡−11213]=\left[\sin^{-1}\dfrac35+\cos^{-1}\dfrac35\right]-\left[\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{12}{13}\right].

Step 4. Apply sin⁡−1t+cos⁡−1t=π2\sin^{-1}t+\cos^{-1}t=\dfrac{\pi}2 to each bracket. =π2−π2=0=\dfrac{\pi}2-\dfrac{\pi}2=0.

✓Final answer

Option (3): 00.

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