Forum

(cosec θ – cot θ)^2...
 
Notifications
Clear all

(cosec θ – cot θ)^2 = (1-cos θ)/(1+cos θ) (ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

1 Posts
2 Users
0 Likes
225 Views
0
Topic starter

Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.

(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

1 Answer
0

(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)

L.H.S. = (cosec θ – cot θ)2

The above equation is in the form of (a - b)2, and expand it

Since (a-b)2 = a2 + b2 – 2ab

Here a = cosec θ and b = cot θ

= (cosec2θ + cot2θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

= (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)

= (1 + cos2θ – 2cos θ)/(1 – cos2θ)

= (1-cos θ)2/(1 – cosθ)(1+cos θ)

= (1-cos θ)/(1+cos θ) = R.H.S.

Therefore, (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)

Hence proved.

(ii)  (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Now, take the L.H.S of the given equation.

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

= [cos2A + (1+sin A)2]/(1+sin A)cos A

= (cos2A + sin2A + 1 + 2sin A)/(1+sin A) cos A

Since cos2A + sin2A = 1, we can write it as

= (1 + 1 + 2sin A)/(1+sin A) cos A

= (2+ 2sin A)/(1+sin A)cos A

= 2(1+sin A)/(1+sin A)cos A

= 2/cos A = 2 sec A = R.H.S.

L.H.S. = R.H.S.

(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Hence proved.

Share:

How Can We Help?