$\frac{1 - tan A}{1 + tan A}$ is equal to

tan (45^{o} - A)

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cot (45^{o} - A)

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tan A

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cot A

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Solution

The correct option is A $tan (45^{o} - A)$
$\frac{1 - t a n A}{1 + t a n A}$
$\Rightarrow \frac{t a n (45^{\circ}) - t a n A}{t a n (45^{\circ}) + t a n A} = \frac{t a n (45) - t a n A}{1 + (t a n 45) (t a n A)}$
Since $t a n (A - B) = \frac{t a n A - t a n B}{1 + t a n A t a n B}$
So, $t a n (45^{\circ} - A) = \frac{t a n 45^{\circ} - t n a A}{1 + (t a n 45^{\circ}) (t a n A)}$
$= \frac{1 - t a n A}{1 + (1) (t a n A)}$
option B is correct.

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Similar questions

1+tanA*tanA/1+cotA*cotA=1-tanA/1-cotA

Prove it

Which of the following options is equal to $\frac{tan A}{1 - cot A} + \frac{cot A}{1 - tan A}$ ?

Which of the following options are equal to $\frac{t a n A}{s e c A - 1} + \frac{t a n A}{s e c A + 1}$ ?

\frac{tan A}{1 - cot A} + \frac{cot A}{1 - tan A} = sec A c o s e c A + 1

tan A / 1- cot A + cot A / 1- tan A = 1+ sec A cosec A.

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