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Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
CBSE | Class 10 | Introduction to trigonometry | Maths | NCERT
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A.

(ii)\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A

Solutions:

(i) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A.

Using the identity function, cosec2A = 1+cot2A, let’s prove the above equation.

L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)

We get by dividing the numerator and denominator by sin A.

= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A

We are aware that cos A/sin A = cot A and 1/sin A = cosec A

= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)

= (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1

= [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)

= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)

=  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)

=  cot A + cosec A = R.H.S.

As a result, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A

Hence Proved.

(ii) \sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A

L.H.S.=\sqrt{\frac{1+\sin A}{1-\sin A}}

L.H.S. numerator and denominator are first divided by cos A,

We are aware that, 1/cos A = sec A and sin A/ cos A = tan A so it becomes,

= √(sec A+ tan A)/(sec A-tan A)

Now, with the help of rationalization, we may arrive at

=\sqrt{\frac{\sec A+\tan A}{\sec A-\tan A}}*\sqrt{\frac{\sec A+\tan A}{\sec A+\tan A}}

=\sqrt{\frac{{{(\sec A+\tan A)}^{2}}}{{{\sec }^{2}}A-{{\tan }^{2}}A}}

= (sec A + tan A)/1

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

As a result, hence proved.

i

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