Question

Cot​ θ + cosec θ -1/ cot θ -cosec θ +1 = 1+cos θ / sin θ

Answer 1

LHS = COTA+COSECA -1 -------------------------------- COTA-COSECA +1 =(COTA+COSECA)-(COSEC^2A- COT^2A) --------------------------------------------- 1-COSECA+COTA = (COTA+COSECA)(1- COSECA+COTA) ----------------------------------- 1-COSECA+COTA =COSECA+COTA =1/SINA + COSA/SINA =(1+COSA) SINA

Answer 2

Answer:

To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

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)2 = (1-cos θ)/(1+cos θ)