Question

Prove that sin^2A/cos^2A + cos^2A/sin^2A = 1/cos^2A*sin^2A - 2

Answer 1

Answer:

prove that:

Sin²A/Cos²A + Cos²A/Sin²A = 1/Cos²A Sin²A - 2

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

Cos

2

A

Sin

2

A

+

Sin

2

A

Cos

2

A

= \begin{lgathered}= \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{lgathered}

=

Cos

2

A.Sin

2

A

Sin

4

A+Cos

4

A

Usinga

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

=

Cos

2

ASin

2

A

1

−2

=RHS

LHS=RHS

Answer 2

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

sin²A + cos²A = 1

Consider the left side

$\frac{sin^2A}{cos^2A}$ + $\frac{cos^2A}{sin^2A}$

= $\frac{sin^2A}{1-sin^2A}$ + $\frac{cos^2A}{1-cos^2A}$

= $\frac{sin^2A(1-cos^2A)+cos^2A(1-sin^2A)}{(1-sin^2A)(1-cos^2A)\\}$

= $\frac{sin^2A-sin^2Acos^2A+cos^2A-sin^2Acos^2A}{1-sin^2A-cos^2A+sin^2Acos^2A}$

= $\frac{sin^2A+cos^2A-2sin^2Acos^2A}{1-(sin^2A+cos^2A)+sin^2Acos^2A}$

= $\frac{1-2sin^2Acos^2A}{sin^2Acos^2A}$

= $\frac{1}{sin^2Acos^2A}$ - $\frac{2sin^2Acos^2A}{sin^2Acos^2A}$

= $\frac{1}{sin^2Acos^2A}$ - 2 = right side ⇒ proven