Prove the identitity- -1-sin-1-sin-tan- -3 MARKS-

Formula: 1 Mark Proof: 2 Marks LHS=√(1 sin θ/1+sin θ)=√((1 sin θ)/(1+sin θ)×(1 sin θ)/(1 sin θ))[Rationalising the denominator] =√((1 sin θ)^2/1 sin^2θ)=√((1 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Relation between Trigonometric Ratios

Prove the ide...

Question

Prove the identitity: $\sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = s e c θ - t a n θ$ [3 MARKS]

Open in App

Solution

Formula: 1 Mark
Proof: 2 Marks

$L H S = \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = \sqrt{\frac{(1 - s i n θ)}{(1 + s i n θ)} \times \frac{(1 - s i n θ)}{(1 - s i n θ})}$ [Rationalising the denominator]
$= \sqrt{\frac{(1 - s i n θ)^{2}}{1 - s i n^{2} θ}} = \sqrt{\frac{(1 - s i n θ)^{2}}{c o s^{2} θ}} = \frac{1 - s i n θ}{c o s θ} = \frac{1}{c o s θ} - \frac{s i n θ}{c o s θ} = s e c θ - t a n θ = R H S$

Suggest Corrections

Similar questions

Q. Prove the ideintity:

\frac{s e c θ + 1 + t a n θ}{s e c θ + 1 - t a n θ} = \frac{1 + s i n θ}{c o s θ}

[4 MARKS]

Q. Prove that

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

[4 MARKS]

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

Q. Prove the following identity:

\sqrt{\frac{1 - sin θ}{1 + sin θ}} = sec θ - tan θ

Join BYJU'S Learning Program

Prove the identitity: √1−sin θ1+sin θ=sec θ−tan θ [3 MARKS]

Formula: 1 Mark Proof: 2 Marks LHS=√1−sin θ1+sin θ=√(1−sin θ)(1+sin θ)×(1− sin θ)(1−sin θ)[Rationalising the denominator] =√(1−sin θ)21−sin2θ=√(1−sin θ)2cos2θ=1−sin θcos θ=1cos θ−sin θcos θ=sec θ−tan θ=RHS

Prove the identitity: $\sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = s e c θ - t a n θ$ [3 MARKS]