Prove the following identities, where the angles involved are acute angles for which the expressions are defined.$$\sqrt{\dfrac{1 +sin\;A}{1 -sin\;A}}=sec\;A +tan\;A$$

Asked by Abhisek | 1 year ago |  158

##### Solution :-

First divide the numerator and denominator of L.H.S. by cos A,

We know that $$\dfrac{1}{cos A}$$ = sec A  and $$\dfrac{sec\; A \;and\; sin \;A}{cos \;A }$$

= tan A and it becomes,

$$\dfrac{\sqrt{(sec A+ tan A)}}{(sec A-tan A)}$$

Now using rationalization, we get

$$\dfrac{ (sec A + tan A)}{1}$$

= sec A + tan A = R.H.S

Hence proved

Answered by Pragya Singh | 1 year ago

### Related Questions

#### Prove the sinθ sin (90° – θ) – cos θ cos (90° – θ) = 0

Prove the sinθ sin (90° – θ) – cos θ cos (90° – θ) = 0

#### Prove that sin 48° sec 48° + cos 48° cosec 42° = 2

Prove that sin 48° sec 48° + cos 48° cosec 42° = 2

#### Prove that tan 20° tan 35° tan 45° tan 55° tan 70° = 1

Prove that tan 20° tan 35° tan 45° tan 55° tan 70° = 1

#### Evaluate the 2sin 3x = \sqrt{3}

Evaluate the $$2sin 3x = \sqrt{3}$$

#### Evalute the following: sin 20°/ cos 70°

Evalute the following:

(i) $$\dfrac{sin 20°}{ cos 70°}$$

(ii) $$\dfrac{ cos 19°}{ sin 71°}$$

(iii) $$\dfrac{sin 21°}{ cos 69°}$$

(iv) $$\dfrac{tan 10°}{ cot 80°}$$

(v) $$\dfrac{sec 11°}{ cosec 79°}$$