Sin 2sin -11-2-tan 2 -1 2- 2cosec-1 4-

The correct option is A 734Let cosec^ 14=θ⇒cosecθ=4 ^2θ=cosec^2θ 1=15 Now, sin^2(sin^ 112)+tan^2(^ 12)+^2(cosec^ 14) =sin^2(π6)+tan^2(π3)+15 =14+3+15=734 ...

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

Byju's Answer

Standard XII

Mathematics

Euler's Representation

sin 2sin -11/...

Question

${sin}^{2} ({sin}^{- 1} \frac{1}{2}) + {tan}^{2} ({sec}^{- 1} 2) + {cot}^{2} ({c o s e c}^{- 1} 4) =$ ______

\frac{73}{4}

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

\frac{37}{2}

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

\frac{89}{4}

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

19

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

Open in App

Solution

The correct option is A $\frac{73}{4}$
Let ${c o s e c}^{- 1} 4 = θ \Rightarrow c o s e c θ = 4$
${cot}^{2} θ = {c o s e c}^{2} θ - 1 = 15$

Now, ${sin}^{2} ({sin}^{- 1} \frac{1}{2}) + {tan}^{2} ({sec}^{- 1} 2) + {cot}^{2} ({c o s e c}^{- 1} 4)$
$= {sin}^{2} (\frac{π}{6}) + {tan}^{2} (\frac{π}{3}) + 15$
$= \frac{1}{4} + 3 + 15 = \frac{73}{4}$

Suggest Corrections

Similar questions

Q. Prove that

{tan}^{2} ({sec}^{- 1} 2) + {cot}^{2} (c o s e c^{- 1} 3) = 11

Q. The value of

t a n^{2} (s e c^{- 1} 3) + c o t^{2} (c o s e c^{- 1} 4)

Q. Find the value of

t a n^{2} (s e c^{- 1} (3))

c o t^{2} (c o s e c^{- 1} (4))

Solve

sin (2 {sin}^{- 1} \frac{4}{5})

∣ ∣ ∣ ∣ \begin{matrix} 10! & 11! & 12! 11! & 12! & 13! 12! & 13! & 14! \end{matrix} ∣ ∣ ∣ ∣ =

__________.

Join BYJU'S Learning Program

sin2(sin−112)+tan2(sec−12)+cot2(cosec−14)=______

The correct option is A 734Let cosec−14=θ⇒cosecθ=4 cot2θ=cosec2θ−1=15 Now, sin2(sin−112)+tan2(sec−12)+cot2(cosec−14) =sin2(π6)+tan2(π3)+15 =14+3+15=734

${sin}^{2} ({sin}^{- 1} \frac{1}{2}) + {tan}^{2} ({sec}^{- 1} 2) + {cot}^{2} ({c o s e c}^{- 1} 4) =$ ______