Cos -1-tan - - sin -1-cot -

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 θ =cos^2 θ/cos θ sin θ sin^2 θ/cos θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identity- 1

cos θ/1-tan θ...

Question

$\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ} =$ ___

$1$

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$cos θ + sin θ$

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$cos θ sin θ$

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$cos θ - sin θ$

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Solution

The correct option is B
$cos θ + sin θ$

$\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ}$

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

$= \frac{cos θ}{\frac{cos θ - sin θ}{cos θ}} + \frac{sin θ}{\frac{sin θ - cos θ}{sin θ}}$

$= \frac{{cos}^{2} θ}{cos θ - sin θ} - \frac{{sin}^{2} θ}{cos θ - sin θ}$

$= \frac{{cos}^{2} θ - {sin}^{2} θ}{cos θ - sin θ}$

$= \frac{(cos θ + sin θ) (cos θ - sin θ)}{cos θ - sin θ} = cos θ + sin θ$

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Similar questions

Evaluate $\frac{1 + t a n θ}{1 + c o t θ}$ , if sin θ = x and cos θ = y.

Q. Solve for

θ

:
i)

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

ii)

\frac{1 + cos θ}{1 - cos θ} = (cosec θ + cot θ)^{2}

Q. If

m s i n θ = n c o s θ

, then show that:

\frac{t a n θ + c o t θ}{t a n θ - c o t θ} = \frac{n s i n θ + m c o s θ}{n s i n θ - m c o s θ} = \frac{n^{2} + m^{2}}{n^{2} - m^{2}}

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cos θ1−tan θ+sin θ1−cot θ= ___

$\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ} =$ ___