Question

Step by step explanation please

if sin theta= a²-b²/a²+b² then cosec theta+cos theta
a) a/a+b b)b+a/b-a. c)a²/a+b. d)a+b/a-b

Answer 1

We have,

sin A = HypotenusePerpendicular=a2+b2a2−b2 

So, we draw a right triangle right angled at B such that

Perpendicular = a2−b2 and, Hypotenuse = a2+b2. and ∠BAC=θ

By Pythagoras theorem, we have

AC2=AB2+BC2

⇒AB2=(a2+b2)2−(a2−b2)2

⇒AB2=(a4+b4+2a2b2)−(a4+b4−2a2b2)

⇒AB2=4a2b2=(2ab)

Answer 2

Answer:

We have,

sin A = HypotenusePerpendicular=a2+b2a2−b2

So, we draw a right triangle right angled at B such that

Perpendicular = a2−b2 and, Hypotenuse = a2+b2. and ∠BAC=θ

By Pythagoras theorem, we have

AC2=AB2+BC2

⇒AB2=(a2+b2)2−(a2−b2)2

⇒AB2=(a4+b4+2a2b2)−(a4+b4−2a2b2)

⇒AB2=4a2b2=(2ab)

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Step-by-step explanation: