Question

50 points !!

solve only if u know
wrong answers will be reported !​

Answer 1

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Answer:

Step-by-step explanation:

tan-1[cosx / (1+sinx)]

=tan-1[ sin(pi/2 - x) / (1+cos(pi/2 - x))]

Let t=pi/2-x

=tan-1[ sint / (1+cost)]

=tan-1[2sin(t/2) cos(t/2) / (2 cos2(t/2))] {since sin2x=2sinx cosx and 1+cos2x=2cos2x }

=tan-1[sin(t/2)/cos(t/2)]

=tan-1[tan(t/2)]

=t/2

=(pi/2 - x)/2

=pi/4 - x/2

Hence Proved.

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Answer 2

Answer:

Heya!

Step-by-step explanation:

Answer:

Step-by-step explanation:

tan-1[cosx / (1+sinx)]

=tan-1[ sin(pi/2 - x) / (1+cos(pi/2 - x))]

Let t=pi/2-x

=tan-1[ sint / (1+cost)]

=tan-1[2sin(t/2) cos(t/2) / (2 cos2(t/2))] {since sin2x=2sinx cosx and 1+cos2x=2cos2x }

=tan-1[sin(t/2)/cos(t/2)]

=tan-1[tan(t/2)]

=t/2

=(pi/2 - x)/2

=pi/4 - x/2

Hope it helps you!