Question

The value of (2+tan^2x+cot^2x)/(secx*cosecx) is equal to?

1. cosx*sinx
2. secx*cosecx
3. cotx
4. tanx

Answer 1

Answer:

2) secx* cosecx

Step-by-step explanation:

(2+ tan^2 X + cot^2 X) / (secx* cosecx)

={2+ (Sec^2 X - 1) +( cosec^2 X- 1) }/ ( secx* cosecx)

={sec^2 X. + cosec^2 X }/ ( secx* cosecx)

=SecX/ Cosecx + cosecx/secx

=tanx + cotx

=

$\frac{ { \sin(x) }^{2} + { \cos(x) }^{2} }{sinx. \cos(x) }$

=cosecx * secx [Ans]

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Answer 2

The value of $\dfrac{2+\tan^2x+\cot^2x}{\sec x\times \csc x}$ is option 2) $\cos x.\csc x.$

Step-by-step explanation:

We have,

$\dfrac{2+\tan^2x+\cot^2x}{\sec x\times \csc x}$

To find, the value of $\dfrac{2+\tan^2x+\cot^2x}{\sec x\times \csc x}=?$

∴ $\dfrac{2+\tan^2x+\cot^2x}{\sec x\times \csc x}$

$=\dfrac{2(\tan x)(\cot x)+\tan^2x+\cot^2x}{\sec x\times \csc x}$

Using identity,

$\tan x.\cot x=1$

$=\dfrac{(\tan x+\cot x)^2}{\sec x\times \csc x}$

[ ∵ $(a+b)^{2} =a^{2}+b^{2} +2ab$]

$=\dfrac{(\tan x+\dfrac{1}{\tan x})^2}{\sec x\times \csc x}$

$=\dfrac{(\dfrac{\tan^2 x+1}{\tan x})^2}{\sec x\times \csc x}$

$=\dfrac{(\dfrac{\sec^2 x}{\tan x})^2}{\sec x\times \csc x}$

Using trigonometric identity,

$\sec^2 A=\tan^2 A+1$

$=\dfrac{\dfrac{\sec^4 x}{\tan^2 x}}{\sec x\times \csc x}$

$=\dfrac{1}{\cos^4 x}\times \dfrac{\cos^2 x}{\sin^2 x} \times \cos x.\sin x$

$=\dfrac{1}{\cos x}\times \dfrac{1}{\sin x}$

$=\cos x.\csc x$

Hence, the value of $\dfrac{2+\tan^2x+\cot^2x}{\sec x\times \csc x}$ is option 2) $\cos x.\csc x.$