Question

Prove
tan Q / Tan ( 90 - Q ) + sin ( 90 - Q ) / cos Q = sec^2Q​

Answer 1

To prove:- tan Q / Tan ( 90° - Q ) + sin ( 90° - Q ) / cos Q = sec^2Q

Step-by-step explanation:

Proof:-

= L. H. S.

{We know

tan(90° - Q) = cot Q

sin (90°- Q) = sin Q}

so,

= Tan Q / cot Q + cos Q / cos Q

=Tan Q / Cot Q + 1

{We know

cot Q = 1 / tan Q}

so,

=tan Q × tan Q / 1+ 1

=tan^ 2 Q + 1

{We know

tan^ 2 Q + 1 = sec^ 2 Q}

so,

=sec^2 Q , hence proved

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