1+tan^2A/1+cot^2A =1+tan^2A/1 + 1/tan^2A =1+tan^2A/tan^2A+1/tan^2A =(1+tan^2A) (tan^2A)/(tan^2A+1) = tan^2A. ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios for Sum of Two Angles

1+tan2A/1+cot...

Question

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

$s e c^{2} A$

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-1

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$c o t^{2} A$

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$t a n^{2} A$

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Solution

The correct option is D
$t a n^{2} A$

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

= $\frac{1 + t a n^{2} A}{1 + \frac{1}{t a n^{2} A}}$

= $\frac{1 + t a n^{2} A}{\frac{t a n^{2} A + 1}{t a n^{2} A}}$

= $\frac{(1 + t a n^{2} A) (t a n^{2} A)}{(t a n^{2} A + 1)}$

= $t a n^{2} A .$

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Similar questions

Q. show that

\frac{1 - {t a n}^{2} A}{{c o t}^{2} A - 1} = {t a n}^{2} A

Q. Prove that

\frac{1 + t a n^{2} A}{1 + c o t^{2} A}

t a n^{2} A .

Prove that:

(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{c o t A})^{2} = t a n^{2} A

(1 + \frac{1}{t a n^{2} A})

(1 + \frac{1}{c o t^{2} A})

\frac{1}{s i n^{2} A s i n^{4} A}

Question 5 (x)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{1 - c o t A})^{2} = t a n^{2} A$

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1+tan2A1+cot2A =

The correct option is D tan2A 1+tan2A1+cot2A =1+tan2A1+1tan2A =1+tan2Atan2A+1tan2A =(1+tan2A)(tan2A)(tan2A+1) = tan2A.

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

The correct option is D
$t a n^{2} A$

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

= $\frac{1 + t a n^{2} A}{1 + \frac{1}{t a n^{2} A}}$

= $\frac{1 + t a n^{2} A}{\frac{t a n^{2} A + 1}{t a n^{2} A}}$

= $\frac{(1 + t a n^{2} A) (t a n^{2} A)}{(t a n^{2} A + 1)}$

= $t a n^{2} A .$