Math, asked by anuragpawar13901, 1 year ago

differentiate log √1-cosmx / √1+cosmx​

Answers

Answered by kritarth24
6

I hope it helps you..mm ..

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anuragpawar13901: but its answer is m cosec(mx)
Answered by Swarup1998
36
\underline{\textsf{Step by step solution :}}

\large{\mathsf{Now,\:\frac{1-cosmx}{1+cosmx}}}

\large{\mathsf{=\frac{(sin^{2}\frac{mx}{2}+cos^{2}\frac{mx}{2})-(cos^{2}\frac{mx}{2}-sin^{2}\frac{mx}{2})}{(sin^{2}\frac{mx}{2}+cos^{2}\frac{mx}{2})+(cos^{2}\frac{mx}{2}-sin^{2}\frac{mx}{2})}}}

\large{\mathsf{=\frac{sin^{2}\frac{mx}{2}+cos^{2}\frac{mx}{2}-cos^{2}\frac{mx}{2}+sin^{2}\frac{mx}{2}}{sin^{2}\frac{mx}{2}+cos^{2}\frac{mx}{2}+cos^{2}\frac{mx}{2}-sin^{2}\frac{mx}{2}}}}

\large{\mathsf{=\frac{2sin^{2}\frac{mx}{2}}{2cos^{2}\frac{mx}{2}}}}

\mathsf{=tan^{2}\frac{mx}{2}}

\to \mathsf{\frac{1-cosmx}{1+cosmx}=tan^{2}\frac{mx}{2}}

\to \mathsf{\sqrt{\frac{1-cosmx}{1+cosmx}}=\sqrt{tan^{2}\frac{mx}{2}}}

\to \mathsf{\sqrt{\frac{1-cosmx}{1+cosmx}}=tan\frac{mx}{2}}

\to \mathsf{log\sqrt{\frac{1-cosmx}{1+cosmx}}=log(tan\frac{mx}{2})}

\small{\textsf{Differentiating both sides w.r. to x, we get}}

\mathsf{\frac{d}{dx}[log\sqrt{\frac{1-cosmx}{1+cosmx}}]=\frac{d}{dx}[log(tan\frac{mx}{2})]}

\mathsf{=\frac{1}{tan\frac{mx}{2}}\frac{d}{dx}(tan\frac{mx}{2})}

\mathsf{=cot\frac{mx}{2}*\frac{m}{2}*sec^{2}\frac{mx}{2}}

\mathsf{=\frac{m}{2}*\frac{cos\frac{mx}{2}}{sin\frac{mx}{2}}*sec^{2}\frac{mx}{2}}

\mathsf{=\frac{m}{2}*\frac{sec\frac{mx}{2}}{sin\frac{mx}{2}}}

\mathsf{=\frac{m}{2sin\frac{mx}{2}cos\frac{mx}{2}}}

\mathsf{=\frac{m}{sinmx}}

\mathsf{=m\:cosecmx}

\to \boxed{\small{\mathsf{\frac{d}{dx}[log\sqrt{\frac{1-cosmx}{1+cosmx}}] = m\:cosecmx}}}

anuragpawar13901: thankyou very much
Swarup1998: thanks for the brainliest :)
Swarup1998: - my pleasure ☺
priya3679: hats off!
priya3679: what an answer!!
Swarup1998: Thank you! ☺
AliceJoy: Cooooooool great brain
Swarup1998: Thanks, Tesla! ☺☺
generalRd: nice use of latex bhaiya
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