The correct option is C 1 We know that,1 + tan^2 θ = sec^2 θ and 1 + cot^2 θ = cosec^2 θ 1/1 + tan^2 θ + 1/1 + cot^2 θ=1/sec^2 θ + 1/cosec^2 θ = cos^2 θ + ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

1/1+tan 2θ+1/...

Question

$\frac{1}{1 + t a n^{2} θ} + \frac{1}{1 + c o t^{2} θ} =$

-1

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Solution

The correct option is C
1

We know that, $1 + t a n^{2} θ = s e c^{2} θ and 1 + c o t^{2} θ = c o s e c^{2} θ$

$\frac{1}{1 + t a n^{2} θ} + \frac{1}{1 + c o t^{2} θ} = \frac{1}{s e c^{2} θ} + \frac{1}{c o s e c^{2} θ}$

$= c o s^{2} θ + s i n^{2} θ$

$= 1$

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Similar questions

Prove:

\frac{{tan}^{2} θ}{1 + {tan}^{2} θ} + \frac{{cot}^{2} θ}{1 + {cot}^{2} θ} = 1

$\frac{t a n^{2} θ}{1 + t a n^{2} θ}$ + $\frac{c o t^{2} θ}{1 + c o t^{2} θ}$ =

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

\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ} =

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

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11+tan2θ+11+cot2θ=

The correct option is C 1 We know that,1+tan2θ=sec2θ and 1+cot2θ=cosec2θ 11+tan2θ+11+cot2θ=1sec2θ+1cosec2θ =cos2θ + sin2θ =1

$\frac{1}{1 + t a n^{2} θ} + \frac{1}{1 + c o t^{2} θ} =$

The correct option is C
1

We know that, $1 + t a n^{2} θ = s e c^{2} θ and 1 + c o t^{2} θ = c o s e c^{2} θ$

$\frac{1}{1 + t a n^{2} θ} + \frac{1}{1 + c o t^{2} θ} = \frac{1}{s e c^{2} θ} + \frac{1}{c o s e c^{2} θ}$

$= c o s^{2} θ + s i n^{2} θ$

$= 1$