Math, asked by deepalisingh803, 1 year ago

Prove that:1+(1-tanA/1-cotA)^2=sec^2A

Answers

Answered by pansumantarkm
9

Step-by-step explanation:

L.H.S. = 1+(\frac{1-tanA}{1-cotA})^{2}

=1+(\frac{1-\frac{1}{cotA} }{1-cotA})^{2}  [∵tanA=\frac{1}{cotA}]

=1+(\frac{\frac{cotA-1}{cotA} }{1-cotA})^{2}

=1+(\frac{-(1-cotA)}{cotA(1-cotA)})^{2}

=1+(\frac{-1}{cotA})^{2}   [∵(-a)^{2}=a^{2}]

=1+\frac{1}{cot^{2}A}    [∵\frac{1}{cotA}=tanA]

=1+tan^{2}A

=sec^{2}A       [∵sec^{2}A-tan^{2}A=1  ⇒sec^{2}A=1+tan^{2}A]

= R.H.S. (Proved)

1+(\frac{1-tanA}{1-cotA}) = sec^{2}A

Hope it will helped you.

Please mark it as Brainliest.

Answered by ayushgautam8687
1

Answer:

I hope it will helps you

Attachments:
Similar questions