MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Chain Rule of Differentiation

Prove that ...

Question

Prove that $\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{s i n^{2} A - c o s^{2} A}$

Open in App

Solution

To prove:- $\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{{sin}^{2} A - {cos}^{2} A}$
Proof: $\frac{(sin A + cos A)^{2} + (sin A - cos A)^{2}}{(sin A - cos A) (sin A + cos A)}$
$= \frac{{sin}^{2} A + {cos}^{2} A + 2 sin A \cdot cos A + {sin}^{2} A + {cos}^{2} A - 2 sin A \cdot cos A}{{sin}^{2} A - {cos}^{2} A}$
$= \frac{1 + 1}{{sin}^{2} A - {cos}^{2} A} = \frac{2}{{sin}^{2} A - {cos}^{2} A} =$ R.H.S.
$[∵ {sin}^{2} A + {cos}^{2} A = 1]$ .

Suggest Corrections

thumbs-up

2

Similar questions

\frac{1 + sin 2 A}{cos 2 A} = \frac{cos A + sin A}{cos A - sin A} = cot (\frac{π}{4} - A)

\frac{1 + cos A}{sin A} = \frac{sin A}{1 - cos A}

\frac{1 + sin A - cos A}{c o s A + sin A - 1} = tan \frac{A}{2}

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

Q. Prove the following identity.

\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{2 s i n^{2} A - 1}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Basic Theorems in Differentiation

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Chain Rule of Differentiation

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon