Question 5 (iv)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(iv) $\frac{(1 + s e c A)}{s e c A} = \frac{s i n^{2} A}{(1 - c o s A)}$
[Hint : Simplify L.H.S and R.H.S separately]

Open in App

Solution

$(i v) \frac{(1 + s e c A)}{s e c A} = \frac{s i n^{2} A}{(1 - c o s A)} L . H . S . = \frac{(1 + s e c A)}{s e c A} = \frac{(1 + \frac{1}{c o s A)}}{\frac{1}{c o s A}} = \frac{\frac{(c o s A + 1)}{c o s A}}{\frac{1}{c o s A}} = c o s A + 1 R . H . S . = \frac{s i n^{2} A}{(1 - c o s A)} = \frac{(1 - c o s^{2} A)}{(1 - c o s A)} = \frac{(1 - c o s A) (1 + c o s A)}{(1 - c o s A)} = c o s A + 1 L . H . S . = R . H . S .$

Suggest Corrections

Similar questions

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

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

(i x) (c o s e c A - s i n A) (s e c A - c o s A) = \frac{1}{(t a n A + c o t A)}

(i x) (c o s e c A - s i n A) (s e c A - c o s A) = \frac{1}{(t a n A + c o t A)}

\frac{c o s A}{(1 + s i n A)} + \frac{(1 + s i n A)}{c o s A} = 2 s e c A

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program