Question

If
$\tan( \alpha ) + \frac{1}{ \tan( \alpha ) } = 2$
then the value of cosec alpha is
a)1
b)1/√2
c)√2
d)√3/2

​

Answer 1

Trigonometric Ratios

The following are the tips and concept that can be use to find the solution:

• Having a basic knowledge of Trigonometric ratios and Angles.
• Trigonometric ratios are sin, cos, tan, cot, sec, cosec.
• The standard angles of these trigonometric ratios are 0°, 30°, 45°, 60° and 90°.

Analyse the values of important angles for all the six trigonometric ratios shown in the table given below:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

[Check the attachment if any case you're not able to see the table]

Let's head to the Question now:

$\implies \tan( \alpha) + \dfrac{1}{\tan( \alpha)} = 2$

$\implies \dfrac{ \tan^{2} ( \alpha) + 1} { \tan( \alpha)} = 2$

$\implies \tan^{2} ( \alpha) + 1 = 2\tan( \alpha)$

$\implies \tan^{2} ( \alpha) + 1 - 2\tan( \alpha) = 0$

$\implies \tan^{2} ( \alpha) - 2\tan( \alpha) + 1 = 0$

$\implies ( \tan( \alpha) - 1)^{2} = 0$

$\implies\tan( \alpha) - 1= 0$

$\implies\tan( \alpha) = 1$

$\implies \boxed{{\bf{tan( \alpha) = {45}^{ \circ}}}}$

Now, we got the angle of $\alpha = {45}^{ \circ}$. Now we can easily find out the value of $\rm{cosec(\alpha)}$.

$\implies \rm{cosec( \alpha) = cosec( {45}^{ \circ})}$

$\implies \boxed{ \bf{{cosec( \alpha) = \sqrt{2}}}}$

Hence, the required value of cosec(α) is √2. So, option (c) is the correct option.

Answer 2

Answer:

Question :-

If $\tan( \alpha ) + \dfrac{1}{ \tan( \alpha ) } = 2$ then the value of cosec alpha is

a)1

b)1/√2

c)√2

d)√3/2

Given :-

• $\tan( \alpha ) + \dfrac{1}{ \tan( \alpha ) } = 2$

Find Out :-

• The value of cosec alpha.

Solution :-

$\leadsto \sf \tan( \alpha ) + \dfrac{1}{ \tan( \alpha ) } = 2$

$\leadsto \sf \dfrac{tan^2(\alpha) + 1}{tan( \alpha )} =\: 2$

$\leadsto \sf tan^2 (α)+1=2tan(α)$

$\leadsto \sf tan^{2} ( \alpha) + 1 - 2 tan( \alpha)= 0$

$\leadsto \sf tan^{2} ( \alpha) - 2 \tan( \alpha) + 1 = 0$

$\leadsto \sf (tan( \alpha) - 1)^{2} = 0$

$\leadsto \sf tan(\alpha) =\: 45^{\circ}$

Now,

➙ cosec(α) = cosec(45°)

➙ cosec(α) = √2

Henceforth, the value of cosec alpha is √2.

Correct options is (c) √2.

[Please refer that attachment for important formula regarding trigonometry]

HOPE IT HELPS YOU :)