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Question : 41 of 100
Marks:
+1,
-0
Solution:
Let
u=tan−1(‌) Put
x=tan‌θ⇒θ=tan−1x, then
u‌‌=tan−1[‌]=tan−1[‌] ‌‌=tan−1[‌]=tan−1[‌] ‌‌=tan−1[‌]=tan−1[tan‌] ⇒‌‌u‌‌=‌=‌tan−1x r⋅tan−1(tan‌θ)=θ] On differentiating both sides w.r.t.
x, we get
‌=‌[∵‌(tan−1x)=‌]...‌ (i) ‌ Also, let
v=sin−1(‌) Put
x=tan‌θ⇒θ=tan−1x, then we get
v=sin−1[‌] ⇒‌v=sin−1[sin‌2‌θ] ⇒‌v=2θ⇒v=2tan−1x ‌ On differentiating both sides w.r.t. ‌x,‌ we get ‌ ‌‌=‌ ‌ Now, ‌‌‌=‌×‌=‌×‌ ‌‌ [from Eqs. (i) and (ii)] ‌ ∴‌‌=‌
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