Question

Cot theta-cos theta/ cot theta+cos theta=cosec theta-1/cosec theta+1

Answer 1

your question is -> prove that (cotθ - cosθ)/(cotθ + cosθ) = (cosecθ - 1)/(cosecθ + 1)

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:

Here we have applied trigonometric formula for solving the given equation.

And we have proved that:

$\dfrac{cot\theta-cos\theta}{cot\theta+cos\theta}$ $= \dfrac{cosec\theta-1}{cosec\theta+1}$

So,

L.H.S = R.H.S

PROVED.

Step-by-step explanation:

L.H.S:

$\dfrac{cot\theta-cos\theta}{cot\theta+cos\theta}$

As we know:

$cot\theta = \dfrac{cos\theta}{sin\theta}$.

Hence substituting the value of $cot\theta$.

$= \dfrac{\frac{cos\theta}{sin\theta}-cos\theta }{\frac{cos\theta}{sin\theta}+cos\theta }$

$= \dfrac{cos\theta-cos\theta\,sin\theta}{cos\theta+cos\theta\,sin\theta}$

$= \dfrac{cos\theta(1-sin\theta)}{cos\theta(1+sin\theta)}$

The formula used below has been given here:

$sin\theta = \dfrac{1}{cosec\theta}$

Therefore substituting the value of $sin\theta$.

$= \dfrac{1-\dfrac{1}{cosec\theta} }{1+\dfrac{1}{cosec\theta} }$

$= \dfrac{cosec\theta-1}{cosec\theta+1}$

= R.H.S

Hence, L.H.S = R.H.S