Cos -1coseccos -1 a for 0- a -1 isA- 1-2-a2B- -3-a2C- 1-2-a2-2-a2

The correct option is A 1√(2 a^2)Let cos^ 1 a=θ⇒cosθ=a Then, cosec θ=1√(1 a^2) Let ^ 1(1√(1 a^2))=ϕ ⇒ϕ=1√(1 a^2) ⇒cosϕ=1√(2 a^2) ∴cos(^ 1(cosec(cos^ 1a)))=1√(2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

cos -1cosecco...

Question

$cos ({cot}^{- 1} (c o s e c ({cos}^{- 1} a)))$ for $0 < a < 1$ is

\frac{1}{\sqrt{2 - a^{2}}}

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

\sqrt{3 - a^{2}}

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

\sqrt{2 - a^{2}}

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

\frac{1}{\sqrt{2 + a^{2}}}

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

Open in App

Solution

The correct option is A $\frac{1}{\sqrt{2 - a^{2}}}$
Let ${cos}^{- 1} a = θ \Rightarrow cos θ = a$
Then, $c o s e c θ = \frac{1}{\sqrt{1 - a^{2}}}$
Let ${cot}^{- 1} (\frac{1}{\sqrt{1 - a^{2}}}) = ϕ$
$\Rightarrow cot ϕ = \frac{1}{\sqrt{1 - a^{2}}}$
$\Rightarrow cos ϕ = \frac{1}{\sqrt{2 - a^{2}}}$
$∴ cos ({cot}^{- 1} (c o s e c ({cos}^{- 1} a))) = \frac{1}{\sqrt{2 - a^{2}}}$

Suggest Corrections

Similar questions

Q. The angle between the tangents from

(α, β)

to the circle

x^{2} + y^{2} = a^{2}

, is

$x^{2} + a^{2} x + b = 0 a n d x^{2} + x + 1 = 0$ have a common roots which of the following is true.

Q. If

\sin^{- 1} (\frac{2 a}{1 - a^{2}}) + \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), where a, x \in (0, 1)

, then, the value of x is

(a) 0
(b)

\frac{a}{2}

\frac{2 a}{1 - a^{2}}

Q. In a

Δ A B C

; the value of

\frac{c o s^{2} \frac{B - C}{2}}{(b + c)^{2}} + \frac{s i n^{2} \frac{B - C}{2}}{(b - c)^{2}}

Q. Simplify the following expressions.

(\frac{a \sqrt{2}}{(1 + a^{2})^{- 1}} - \frac{2 \sqrt{2}}{a^{- 1}}) \frac{a^{- 3}}{1 - a^{- 2}}

Join BYJU'S Learning Program

cos(cot−1(cosec(cos−1a))) for 0<a<1 is

The correct option is A 1√2−a2Let cos−1a=θ⇒cosθ=a Then, cosec θ=1√1−a2 Let cot−1(1√1−a2)=ϕ ⇒cotϕ=1√1−a2 ⇒cosϕ=1√2−a2 ∴cos(cot−1(cosec(cos−1a)))=1√2−a2

$cos ({cot}^{- 1} (c o s e c ({cos}^{- 1} a)))$ for $0 < a < 1$ is