Question

y=tan^-1{(sinx)/(1+cosx)}

Answer 1

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Notice,

tan−1(sinx1+cosx)=tan−1(2sinx2cosx21+2cosx2−1)tan−1⁡(sin⁡x1+cos⁡x)=tan−1⁡(2sin⁡x2cos⁡x21+2cos⁡x2−1)

=tan−1(2sinx2cosx22cos2x2)=tan−1⁡(2sin⁡x2cos⁡x22cos2⁡x2)

=tan−1(sinx2cosx2)=tan−1⁡(sin⁡x2cos⁡x2)

=tan−1(tanx2)=x2=tan−1⁡(tan⁡x2)=x2

and,

tan−1(cosx1+sinx)=tan−1⎛⎝⎜⎜1−tan2x21+tan2x21+2tanx21+tan2x2⎞⎠⎟⎟tan−1⁡(cos⁡x1+sin⁡x)=tan−1⁡(1−tan2⁡x21+tan2⁡x21+2tan⁡x21+tan2⁡x2)

=tan−1(1−tan2x21+tan2x2+2tanx2)=tan−1⁡(1−tan2⁡x21+tan2⁡x2+2tan⁡x2)

=tan−1((1−tanx2)(1+tanx2)(1+tanx2)2)=tan−1⁡((1−tan⁡x2)(1+tan⁡x2)(1+tan⁡x2)2)

=tan−1(1−tanx21+tanx2)=tan−1⁡(1−tan⁡x21+tan⁡x2)

=tan−1(tanπ4−tanx21+tanπ4tanx2)=tan−1⁡(tan⁡π4−tan⁡x21+tan⁡π4tan⁡x2)

=tan−1(tan(π4−x2))=tan−1⁡(tan⁡(π4−x2))

=π4−x2=π4−x2

hence,

ddx(sinx1+cosx)ddx(cosx1+sinx)ddx(sin⁡x1+cos⁡x)ddx(cos⁡x1+sin⁡x)

=ddx(x2)ddx(π4−x2)=ddx(x2)ddx(π4−x2)

=1/2−1/2=1/2−1/2

=−1

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