Question

Prove this particular question.​

Answer 1

Given :

$\\ :\normalsize\boxed{\bf\dfrac{1}{1+\tan^2A}+\dfrac{1}{1+\cot^2A}=1}$

To Find :

Prove.

Solution :

Analysis :

Here first we have to solve the LHS using the required identity. Then after getting the LHS we can prove it.

Explanation :

$\\ \normalsize\sf\dfrac{1}{1+\tan^2A}+\dfrac{1}{1+\cot^2A}=1$

Now the LHS :

$\\ :\implies\normalsize\sf\dfrac{1}{1+\tan^2A}+\dfrac{1}{1+\cot^2A}$

$\\ \bf Using\ \tan^2A=\dfrac{\sin^2A}{\cos^2A},$

$\\ :\implies\normalsize\sf\dfrac{1}{1+\dfrac{\sin^2A}{cos^2A}}+\dfrac{1}{1+\cot^2A}$

$\\ \bf Using\ \cot^2A=\dfrac{\cos^2A}{\sin^2A},$

$\\ :\implies\normalsize\sf\dfrac{1}{1+\dfrac{\sin^2A}{\cos^2A}}+\dfrac{1}{1+\dfrac{\cos^2A}{\sin^2A}}$

Taking LCM of cos²A and 1 = cos²A; Taking LCM of sin²A and 1 = sin²A,

$\\ :\implies\normalsize\sf\dfrac{1}{\dfrac{\cos^2A+\sin^2A}{\cos^2A}}+\dfrac{1}{\dfrac{\sin^2A+\cos^2A}{\sin^2A}}$

Doing a reciprocal,

$\\ :\implies\normalsize\sf1\times\dfrac{\cos^2A}{\cos^2A+\sin^2A}+1\times\dfrac{\sin^2A}{\sin^2A+\cos^2A}$

Multiplying cos²A and sin²A by 1,

$\\ :\implies\normalsize\sf\dfrac{1\times\cos^2A}{\cos^2A+\sin^2A}+\dfrac{1\times\sin^2A}{\sin^2A+\cos^2A}$

$\\ :\implies\normalsize\sf\dfrac{\cos^2A}{\cos^2A+\sin^2A}+\dfrac{\sin^2A}{\sin^2A+\cos^2A}$

Arranging (sin²A + cos²A) as (cos²A + sin²A),

$\\ :\implies\normalsize\sf\dfrac{\cos^2A}{\cos^2A+\sin^2A}+\dfrac{\sin^2A}{\cos^2A+\sin^2A}$

Doing LCM,

$\\ :\implies\normalsize\sf\dfrac{\cos^2A+\sin^2A}{\cos^2A+\sin^2A}$

Cancelling the same terms,

$\\ :\implies\normalsize\sf\dfrac{\cancel{\cos^2A+\sin^2A}}{\cancel{\cos^2A+\sin^2A}}$

$\\ \normalsize\therefore\boxed{\bf LHS=1.}$

Here,

⇒ 1(LHS) = 1(RHS)

∴ LHS = RHS.

• Hence proved.

$\\ \normalsize\therefore\boxed{\bf\dfrac{1}{1+\tan^2A}+\dfrac{1}{1+\cot^2A}=1\ \ (proved).}$