Question

(1+sin theta/ cos theta)+(cos theta/1+sin theta)= 2 sec theta​

Answer 1

Answer:

1+sin theta/cos theta +cos theta/1+sin theta

1+sin theta/cos theta +cos theta/1+sin theta={(1+sin^2 theta)+cos^2 theta}/cos theta (1+sin theta)

1+sin theta/cos theta +cos theta/1+sin theta={(1+sin^2 theta)+cos^2 theta}/cos theta (1+sin theta)={1+sin^2 theta+2 sin theta+cos^2 theta}/cos theta (1+sin theta)

1+sin theta/cos theta +cos theta/1+sin theta={(1+sin^2 theta)+cos^2 theta}/cos theta (1+sin theta)={1+sin^2 theta+2 sin theta+cos^2 theta}/cos theta (1+sin theta)=2(1+sin theta)/cos theta (1+sin theta)

1+sin theta/cos theta +cos theta/1+sin theta={(1+sin^2 theta)+cos^2 theta}/cos theta (1+sin theta)={1+sin^2 theta+2 sin theta+cos^2 theta}/cos theta (1+sin theta)=2(1+sin theta)/cos theta (1+sin theta)=2/ cos theta

1+sin theta/cos theta +cos theta/1+sin theta={(1+sin^2 theta)+cos^2 theta}/cos theta (1+sin theta)={1+sin^2 theta+2 sin theta+cos^2 theta}/cos theta (1+sin theta)=2(1+sin theta)/cos theta (1+sin theta)=2/ cos theta=2 sec theta [Proved.]

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Answer 2

LHS

=1+sinθ/cosθ +cosθ/1+sinθ

={(1+sin²θ)+cos²θ}/cosθ (1+sinθ)

={1+sin²θ+2 sinθ+cos²θ}/cosθ (1+sinθ)

=2(1+sinθ)/cosθ (1+sinθ)

=2/ cosθ

=2 secθ= RHS.

[Proved.]