If 3 -4- then- -2 -1-sin 2- is equal to

The correct option is C 1 cot α 1 cot α We have: =√(2 cot α +1/sin^2 α) =√(2cos α/sin α+1/sin^2 α) =√(2 sin α cos α +sin^2 α+cos^2 α/sin^2 α) =√((sin α +cos ...

You visited us 97 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

If 3 π/4<α<π ...

Question

If $\frac{3 π}{4} < α < π$ then, $\sqrt{2 c o t α + \frac{1}{s i n^{2} α}}$ is equal to

$1 - c o t α$

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

$1 + c o t α$

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

$- 1 + c o t α$

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

$- 1 - c o t α$

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

Open in App

Solution

The correct option is D
$- 1 - c o t α$

$- 1 - c o t α$

We have:

$= \sqrt{2 c o t α + \frac{1}{s i n^{2} α}}$

$= \sqrt{\frac{2 c o s α}{s i n α} + \frac{1}{s i n^{2} α}}$

$= \sqrt{\frac{2 s i n α c o s α + s i n^{2} α + c o s^{2} α}{s i n^{2} α}}$

$= \sqrt{\frac{(s i n α + c o s α)^{2}}{s i n^{2} α}}$

$= \sqrt{(1 + c o t α)^{2}}$

$= | 1 + c o t α |$

$= - (1 + c o t α) [W h e n \frac{3 π}{4} < α < π, c o t α < - 1 \Rightarrow c o t α + 1 < 0]$

$= - 1 - c o t α$

Suggest Corrections

Similar questions

If $c o s A = \frac{4}{5}$ and $c o s B = \frac{12}{13}, \frac{3 π}{2} < A, B < 2 π$ , find the value of the following

$(i) c o s (A + b)$

$(i i) s i n (A - B)$

Q. If

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 1, w h e n 0 < x \leq \frac{3 π}{4} 2 s i n \frac{2}{9} x, w h e n \frac{3 π}{4} < x < π \end{matrix}

, then

A pole of height 12 feet is reduced with a scale

factor of \frac{3}{4} to form a new pole. Find the height of

the new pole.

\frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}

will be purely imaginary, if

θ

Q. For any angle

A, \sqrt{1 + s i n A} = ∣ ∣ c o s \frac{A}{2} + s i n \frac{A}{2} ∣ ∣, \sqrt{1 - s i n A} = ∣ ∣ c o s \frac{A}{2} - s i n \frac{A}{2} ∣ ∣

Use this information to answer the following.

\sqrt{1 + s i n A} - \sqrt{1 - s i n A} = - 2 c o s \frac{A}{2} \Rightarrow

Join BYJU'S Learning Program

If 3π4<α<π then, √2cotα+1sin2α is equal to

The correct option is D −1−cotα −1−cotα We have: =√2cotα+1sin2α =√2cosαsinα+1sin2α =√2sinαcosα+sin2α+cos2αsin2α =√(sinα+cosα)2sin2α =√(1+cotα)2 =|1+cotα| =−(1+cotα)[When3π4<α<π,cotα<−1⇒cotα+1<0] =−1−cotα

If $\frac{3 π}{4} < α < π$ then, $\sqrt{2 c o t α + \frac{1}{s i n^{2} α}}$ is equal to