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Question 148 of 177

Q.Prove that: sin⁡−1(35)+cos⁡−1(1213)=sin⁡−1(5665)\sin^{-1}\left(\dfrac{3}{5}\right) + \cos^{-1}\left(\dfrac{12}{13}\right) = \sin^{-1}\left(\dfrac{56}{65}\right).

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 4mImportance★★★★★
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Let A=sin⁡−1(3/5)A=\sin^{-1}(3/5), B=cos⁡−1(12/13)B=\cos^{-1}(12/13); compute sin⁡(A+B)\sin(A+B) using the addition formula and check the range.

Let A=sin⁡−1(35)A = \sin^{-1}\left(\dfrac{3}{5}\right), so sin⁡A=35\sin A = \dfrac35, and since A∈[−π2,π2]A\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] with sin⁡A>0\sin A>0, AA is acute, giving cos⁡A=45\cos A = \dfrac45.

Let B=cos⁡−1(1213)B = \cos^{-1}\left(\dfrac{12}{13}\right), so cos⁡B=1213\cos B = \dfrac{12}{13}, and since B∈[0,π]B\in[0,\pi] with cos⁡B>0\cos B>0, BB is acute, giving sin⁡B=513\sin B = \dfrac{5}{13}.

Using sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B+\cos A\sin B:

sin⁡(A+B)=35⋅1213+45⋅513=3665+2065=5665\sin(A+B) = \frac35\cdot\frac{12}{13}+\frac45\cdot\frac{5}{13} = \frac{36}{65}+\frac{20}{65} = \frac{56}{65}

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