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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

1-tan245o - A...

Question

$\frac{1 - t a n^{2} (45^{o} - A)}{1 + t a n^{2} (45^{o} - A)} =$

s i n 2 A

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c o s 2 A

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t a n 2 A

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c o t 2 A

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Solution

The correct option is A $s i n 2 A$
$= \frac{1 - {tan}^{2} (45^{o} - A)}{1 + {tan}^{2} (45^{o} - A)}$ , we know
$cos 2 A = \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A}$
$= cos 2 (45 - A)$
$= cos (90 - 2 A)$
$= sin 2 A$

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Similar questions

A = 30^{0}

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

30^{\circ}

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

$\frac{sin 2 A}{1 + cos 2 A}$ =

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $s i n^{2} A - c o s^{2} A$

3.tan A = $\frac{2 t a n \frac{A}{2}}{1 - t a n^{2} \frac{A}{2}}$

4. $s i n^{2} \frac{A}{2}$ = 1 - cosA

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

6. $s i n^{2} A$ = $\frac{2 t a n A}{1 - t a n^{2} A}$

If $3 c o t A = 4 . Prove that \frac{1 - t a n^{2} A}{1 + t a n^{2} A} = c o s^{2} A - s i n^{2} A$ .

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