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Question : 25 of 160
Marks:
+1,
-0
Solution:
‌∵tanh−1(x)=‌‌ln(‌)(‌ for ‌|x|<1) ‌‌‌‌‌‌ so, ‌tanh−1(sin‌θ)‌=‌‌ln(‌) ‌‌=‌| (1+sin‌θ)2 |
| (1−sin‌2θ) |
=(‌)2 ‌=‌‌ln(‌)2 ‌=ln(‌)‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) ‌ and ‌cosh−1(x)‌=ln(x+√x2−1) ∴cosh−1( secθ)‌=ln(‌+√‌) ‌=ln(‌)‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) By comparing Eqs. (i) and (ii), we get
tanh−1(sin‌θ)=cosh−1( secθ)
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