MyQuestionIcon

4

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Sign of Trigonometric Ratios in Different Quadrants

Solve: sins...

Question

Solve: $sin {{sin}^{- 1} (\frac{1 - \sqrt{3}}{2 \sqrt{2}}) + {cos}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$

A

\frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

\frac{\sqrt{3}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

\frac{1}{\sqrt{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0$
$sin {{sin}^{- 1} (\frac{1 - \sqrt{3}}{2 \sqrt{2}}) + {cos}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$
$= sin {- {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + {cos}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$ [Since ${sin}^{- 1} (- x) = - {sin}^{- 1} x$ ]
$= sin {- {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + \frac{π}{2} - {sin}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$
$= sin {\frac{π}{2} - {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) - {sin}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$
$= sin {\frac{π}{2} - \frac{π}{2}} = 0$

$∵ [{sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + {sin}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}}) = {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}} . \frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{\sqrt{3} + 1}{2 \sqrt{2}} . \frac{\sqrt{3} + 1}{2 \sqrt{2}}) = {sin}^{- 1}]$

Suggest Corrections

thumbs-up

0

Similar questions

0^{\circ} < θ < 90^{\circ}

2 cos 3 θ = 1

2 sin 2 θ = \sqrt{3}

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

Q. Solve the following equations for x:

(i)

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1}

\sin

x

=

\tan

^{$-$ 1} (2

\sin

x

x \neq \frac{π}{2}

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

$Solve the following equations : (i) s i n^{2} θ - c o s θ = \frac{1}{4} (i i) 2 c o s^{2} θ - 5 c o s θ + 2 = 0 (i i i) 2 s i n^{2} x + \sqrt{3} c o s x + 1 = 0 (i v) 4 s i n^{2} θ - 8 c o s θ + 1 = 0 (v) t a n^{2} x + (1 - \sqrt{3}) t a n x - \sqrt{3} = 0 (v i) 3 c o s^{2} θ - 2 \sqrt{3} s i n θ c o s θ - 3 s i n^{2} θ = 0 (v i i) c o s 4 θ = c o s 2 θ$

2 {cos}^{2} θ - \sqrt{3} sin θ + 1 = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Trigonometric Functions in a Unit Circle

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Sign of Trigonometric Ratios in Different Quadrants

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon