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

1 Answer

Solution :-

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

Ncert solutions class 10 chapter 8-12

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

Ncert solutions class 10 chapter 8-13

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

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

Hence proved

Answered by Pragya Singh | 1 year ago

Related Questions

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) \( \dfrac{sec 11°}{ cosec 79°}\)

Class 10 Maths Introduction to Trigonometry View Answer