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INVERSE TRIGNOMETRIC FUNCTIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINVERSE TRIGNOMETRIC FUNCTIONS3 Marks
Question
Evaluate:
$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)$
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Answer

$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)$
$=\cos\Bigg[\sin^{-1}\Bigg(\frac{3}{5}\sqrt{1-\Big(\frac{5}{13}\Big)^2}+\frac{5}{13}\sqrt{1-\Big(\frac{3}{5}\Big)^2}\Bigg)\Bigg]$
$\Big\{\text{Since }\sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\Big[\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]\Big\}$
$=\cos\Big[\sin^{-1}\Big(\frac{3}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{4}{5}\Big)\Big]$
$=\cos\Big[\sin^{-1}\Big(\frac{56}{65}\Big)\Big]$
$=\cos\Bigg[\cos^{-1}\Bigg(\sqrt{1-\Big(\frac{56}{65}\Big)^2}\Bigg]$
$\Big\{\text{Since }\sin^{-1}\text{x}=\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big)\Big\}$
$=\cos\Big[\cos^{-1}\Big(\frac{33}{65}\Big)\Big]$
$=\frac{33}{65}$ $\Big\{\text{Since }\cos\big(\cos^{-1}\text{x}\big)=\text{x}\text{ as }\text{x}\in[0,1]\Big\}$
Hence,
$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)=\frac{33}{65}$

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