Question

(iv); (1+cot theta+cosec theta)(1+cot theta-cosec theta ) =(cot theta/ 2 -tan theta/ 2 )​

Answer 1

Answer:

Step-by-step explanation:

LHS = (cotθ - cosθ)/(cotθ + cosθ)

putting, cotθ = cosθ/sinθ

= (cosθ/sinθ - cosθ)/(cosθ/sinθ + cosθ)

= {cosθ(1/sinθ - 1)}/{cosθ(1/sinθ - 1)}

= (1/sinθ - 1)/(1/sinθ - 1)

we know, 1/sinθ = cosecθ

= (cosecθ - 1)/(cosecθ + 1) = RHS

hence proved

Answer 2

Answer:

Let a = 1+CotX

and b = CosecX

so We can write to LHS of above given equation as (a+b) (a-b) = (1+CotX)^2 - CosecX^2 = 2CotX

RHS = CotX/2 - tanX/2

= [1-tanX/2 ^2]/ tan X/2

= multiplying 2 at numerator and denominator and comparing with formula Cot2X = (1-tanX ^2)/2tanX we get 2 CotX

Thus LHS= RHS