Prove the following trigonometric identities- 1-sin-1-sin-sec-tan-2

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

Trigonometric Identities

Prove the fol...

Question

Prove the following trigonometric identities:

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

Open in App

Solution

Suggest Corrections

Similar questions

Prove the following trigonometric identities:

$\frac{1 - c o s θ}{s i n θ} = \frac{s i n θ}{1 + c o s θ}$

\frac{1 - sin θ}{1 + sin θ} = {(sec θ - tan θ)}^{2}

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

Prove the following

1. $(1 - {sin}^{2} A) s e c^{2} A = 1$

2. ${sec}^{4} θ - {sec}^{2} θ = {tan}^{4} θ + {tan}^{2}$

3. $(sec θ - tan θ)^{2} = \frac{1 - sin θ}{1 + sin θ}$

4. $\frac{tan θ + sec θ - 1}{tan θ - sec θ +} = \frac{1 + sin θ}{cos θ}$

\frac{1 - sin θ}{1 + sin θ} = {(sec θ - tan θ)}^{2}

Join BYJU'S Learning Program