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Trigonometric Ratios of Multiples of an Angle

If y = √1 -...

Question

If $y = \frac{\sqrt{1 - sin 4 a} + 1}{\sqrt{1 + sin 4 A} - 1}$ , then one of the values of $y$ is

A

- tan A

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B

cot A

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C

tan (\frac{π}{4} + A)

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D

- cot (\frac{π}{4} + A)

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Solution

The correct options are
A $- tan A$
B $cot A$
C $tan (\frac{π}{4} + A)$
D $- cot (\frac{π}{4} + A)$
$y = \frac{\sqrt{(cos 2 A - sin 2 A)^{2}} + 1}{\sqrt{(cos 2 A + sin 2 A)^{2}} - 1} \Rightarrow y = \frac{\pm (cos 2 A - sin 2 A) + 1}{\pm (cos 2 A + sin 2 A) - 1}$
So $y_{1}, y_{2}, y_{3}, y_{4}$ are the four values of $y$
Then
$y_{1} = \frac{cos 2 A - sin 2 A + 1}{cos 2 A + sin 2 A - 1} = \frac{(1 + cos 2 A) - sin 2 A}{(cos 2 A - 1) + sin 2 A}$
$= \frac{2 {cos}^{2} A - 2 sin A cos A}{- 2 {cos}^{2} A + 2 sin A cos A} = \frac{cos A (cos A - sin A)}{sin A (cos A - sin A)} = cot A$
$y_{2} = \frac{- (cos 2 A - sin 2 A) + 1}{- (cos 2 A - sin 2 A) - 1} = \frac{(1 + cos 2 A) + sin 2 A}{- (cos 2 A - 1) - sin 2 A}$
$= \frac{2 {sin}^{2} A + 2 sin A cos A}{- 2 {cos}^{2} A - 2 sin A cos A} = - tan A$
$y_{3} = \frac{(cos 2 A - sin 2 A) + 1}{- (cos 2 A - sin 2 A) - 1} = \frac{(1 + cos 2 A) - sin 2 A}{- (cos 2 A - 1) - sin 2 A}$
$= \frac{2 {cos}^{2} A - 2 sin A cos A}{- 2 {cos}^{2} A - 2 sin A cos A} = - \frac{cos A - sin A}{cos A + sin A} = - \frac{1 - tan A}{1 + tan A}$
$= - tan (\frac{π}{4} - A) = - cot (\frac{π}{4} + A)$
$y_{4} = \frac{- (cos 2 A - sin 2 A) + 1}{(cos 2 A - sin 2 A) - 1} = \frac{(1 + cos 2 A) + sin 2 A}{- (cos 2 A - 1) + sin 2 A}$
$= \frac{2 {sin}^{2} A + 2 sin A cos A}{- 2 {sin}^{2} A + 2 sin A cos A} = \frac{cos A + sin A}{cos A - sin A} = \frac{1 + tan A}{1 - tan A} = tan (\frac{π}{4} + A)$
Hence, all the options are correct.

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Similar questions

Q. If cos A + cos² A = 1, then (sin² A + sin⁴ A) = ?
(a) 1
(b) 2
(c) 3
(d) 4

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A + B + C = π

{sin}^{4} A + {sin}^{4} B + {sin}^{4} C = \frac{3}{2} + 2 cos A cos B cos C + \frac{1}{2} cos 2 A cos 2 B cos 2 C

\frac{{cos}^{2} A}{{cos}^{2} B} + \frac{{sin}^{4} A}{{sin}^{2} B} = 1

, then show that

\frac{{cos}^{2} A}{{cos}^{2} B} + \frac{{sin}^{4} A}{{sin}^{2} B} = 1

$I f c o s^{4} A - s i n^{4} A = \frac{3}{5}, t h e n 2 c o s^{2} A - 1$ = ?

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