CBSE 12th Standard Maths Subject Inverse Trigonometric Functions HOT Questions 4 Mark Questions 2021
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CBSE 12th Standard Maths Subject Inverse Trigonometric Functions HOT Questions 4 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Maths
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Find the values of each of the expressions in Exercises : \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
(a)
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CBSE 12th Standard Maths Subject Inverse Trigonometric Functions HOT Questions 4 Mark Questions 2021 Answer Keys
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Given expression \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
Putting \(\sin ^{-1}\left(\frac{3}{5}\right)=x \text { and } \cot ^{-1}\left(\frac{3}{2}\right)=y\)
Or \( \sin (x)=3 / 5 \ and \ \cot y=3 / 2 \)
\(Now, \sin (x)=3 / 5 \Rightarrow \cos x=\sqrt{1-\sin ^2 x}=4 / 5\ and \ \sec x=5 / 4 \)
(Using identities \(\cos \mathrm{x}=\sqrt{1-\sin ^2 x} \text { and } \sec \mathrm{x}=1 / \cos\))
\(\tan x=\sqrt{\sec ^2 x-1}=\sqrt{\frac{25}{16}-1}=3 / 4 \text { and } \tan y=1 / \cot (y)=2 / 3\)
we can written as
\( \tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\cot ^{-1} \frac{3}{2}\right)=\tan (x+y) \)
\(=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}} \)
\(=17 / 6\)
