The correct option is A cosθ+sinθ cos θ/1 tan θ + sin θ/1 cot θ = cos θ/1 sin θ/cos θ + sin θ/1 cos θ/sin θ = cos θ/cos θ sin θ/cos θ + sin θ/sin θ cos θ/sin ...

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

Trigonometric Identities

cosθ/1-tanθθ+...

Question

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

$c o s θ + s i n θ$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$c o s θ - s i n θ$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$c o s θ . s i n θ$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$c o s θ + s i n θ$

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

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

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

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

= $\frac{c o s^{2} θ}{c o s θ - s i n θ} - \frac{s i n^{2} θ}{c o s θ - s i n θ}$

= $\frac{c o s^{2} θ - s i n^{2} θ}{c o s θ - s i n θ}$

= $\frac{(c o s θ + s i n θ) (c o s θ - s i n θ)}{c o s θ - s i n θ}$ ..... $a^{2} - b^{2} = (a + b) (a - b)$

= $c o s θ + s i n θ$

Suggest Corrections

Similar questions

Q. Prove the following.

(1) secθ (1 – sinθ) (secθ + tanθ) = 1

(2) (secθ + tanθ) (1 – sinθ) = cosθ

(3) sec²θ + cosec²θ = sec²θ × cosec²θ

(4) cot²θ – tan²θ = cosec²θ – sec²θ

(5) tan⁴θ + tan²θ = sec⁴θ – sec²θ

(6)

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

(7) sec⁶x – tan⁶x = 1 + 3sec²x × tan²x

(8)

\frac{\tan θ}{s e c θ + 1} = \frac{s e c θ - 1}{\tan θ}

(9)

\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ

(10)

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

Q. Solve

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

Q. Prove that

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

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

Join BYJU'S Learning Program

cosθ1−tanθ+sinθ1−cotθ = ___

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