Math, asked by AnswerMe314, 3 months ago

Help with it pls, best answer is brainliest.​

Attachments:

Answers

Answered by muralipriya2015007
0

Step-by-step explanation:

LHS = \frac{Sin^2A}{Cos^2A} + \frac{Cos^2A}{Sin^2A}

Cos

2

A

Sin

2

A

+

Sin

2

A

Cos

2

A

= \begin{gathered}= \frac{Sin^4A + Cos^4A}{Cos^2A . Sin^2A}\\\\Using\: a^2 + b^2 = (a+b)^2 - 2ab\\\\a = Cos^2A \: \& \:b = Sin^2A\\\\= \frac{(Sin^2A + Cos^2A)^2 - 2Sin^2A Cos^2A}{Cos^2A Sin^2A} \\\\Sin^2A + Cos^2A = 1\\\\= \frac{1 -2Sin^2A Cos^2A}{Cos^2A Sin^2A}\end{gathered}

=

Cos

2

A.Sin

2

A

Sin

4

A+Cos

4

A

Using a 2 +b 2 =(a+b) 2

−2ab

a=Cos

2

A&b=Sin

2

A

=

Cos

2

ASin

2

A

(Sin

2

A+Cos

2

A)

2

−2Sin

2

ACos

2

A

Sin

2

A+Cos

2

A=1

=

Cos

2

ASin

2

A

1−2Sin

2

ACos

2

A

\begin{gathered}= \frac{1}{Cos^2A Sin^2A} - 2\\\\= RHS\end{gathered}

=

Cos

2

ASin

2

A

1

−2

=RHS

Similar questions