Given that, sinθ1 cosθ = cos ecθ #160; + θ Now, L.H.S = sinθ( 1 + cosθ)sin^2θ = cos ecθ #160; + θ =R.H.S proved. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relation between Trigonometric Ratios

Prove : sinθ...

Question

Prove :
$\frac{sin θ}{1 - cos θ} = csc θ + cot θ$

Open in App

Solution

Given that,
$\frac{sin θ}{1 - cos θ} = cos e c θ + cot θ$
Now,
$L . H . S = \frac{sin θ (1 + cos θ)}{{sin}^{2} θ}$
$= cos e c θ + cot θ$
$= R . H . S$ proved.

Suggest Corrections

Similar questions

\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = csc θ + cot θ

Prove the following trigonometric identities:

Prove that:

(i) $\sqrt{\frac{s e c θ - 1}{s e c θ + 1}} + \sqrt{\frac{s e c θ + 1}{s e c θ - 1}} = 2 c o s e c θ$

(ii) $\sqrt{\frac{1 + s i n θ}{1 - s i n θ}} + \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = 2 c o s e c θ$

(iii) $\sqrt{\frac{1 + c o s θ}{1 - c o s θ}} + \sqrt{\frac{1 - c o s θ}{1 + c o s θ}} = 2 c o s e c θ$

(iv) $\frac{s e c θ - 1}{s e c θ + 1} = {(\frac{s i n θ}{1 + c o s θ})}^{2}$

Join BYJU'S Learning Program