The correct option is D α√(1 α^2) According to question, sinθ=α , #160; then, #160; cosθ=√(1 α^2) #160; tanθ=sinθcosθ #160; t ...

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

Range of Trigonometric Ratios from 0 to 90 Degrees

If sinθ=α t...

Question

If $sin θ = α$ then, $tan θ$ =......... .

α

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

\sqrt{1 - α^{2}}

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

\frac{α}{1 - α^{2}}

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

\frac{α}{\sqrt{1 - α^{2}}}

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

Open in App

Solution

The correct option is D $\frac{α}{\sqrt{1 - α^{2}}}$
According to question,

$sin θ = α$ ,

then, $cos θ = \sqrt{1 - α^{2}}$

$⟹ tan θ = \frac{sin θ}{cos θ}$

$⟹ tan θ = \frac{α}{\sqrt{1 - α^{2}}}$

Suggest Corrections

Similar questions

α

⎡ ⎢ ⎣ \begin{matrix} 1 & α & α^{2} α & 1 & α α^{2} & α & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦

1 + α + α^{2} = ?

α

⎡ ⎢ ⎣ \begin{matrix} 1 & α & α^{2} α & 1 & α α^{2} & α & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦

1 + α + α^{2} =

α

⎡ ⎢ ⎣ \begin{matrix} 1 & α & α^{2} α & 1 & α α^{2} & α & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦

1 + α + α^{2} =

1, α, α^{2} . . . . . . α^{n}

n^{t h}

^{n} C_{1} +^{n} C_{2} . α +^{n} C_{3} . α^{2} . . . . . . . . +^{n} C_{n} . α^{n}

α

α = \sqrt[5]{1}

2^{| 1 + α + α^{2} + α^{3} + α^{- 1} |}

Join BYJU'S Learning Program