Question

If sec theta=3/2,then find the value of 2 cos theta-tan^2 theta/sin^2theta +1​

Answer 1

Answer:

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Answer 2

Answer:

The value of the given trigonometric expression is

$\frac{2cos\theta-tan^2\theta}{sin^2\theta+1}=\frac{3}{56}$

Step-by-step explanation:

Given that,

$sec\theta=\frac{3}{2}$

We are required to find the value of the trigonometric expression which is mentioned below.

$\frac{2cos\theta-tan^2\theta}{sin^2\theta+1}$

If $sec\theta=\frac{3}{2}$ then

$cos\theta=\frac{1}{sec\theta}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$

This indicates that the adjacent side of the triangle is 2 units and the hypotenuse is 3 units. Hence, the opposite side can be found as follows-

$hyp^2=opp^2+adj^2\\3^2=opp^2+2^2\\= > opp=\sqrt{9-4}=\sqrt{5}\:units$

Therefore, the other trigonometric ratios are

$sin\theta=\frac{opp}{hyp}=\frac{\sqrt{5}}{3}\\tan\theta=\frac{opp}{adj}=\frac{\sqrt{5}}{2}$

Hence our trigonometric expression becomes

$\frac{2cos\theta-tan^2\theta}{sin^2\theta+1}=\frac{2\frac{2}{3}-(\frac{\sqrt{5}}{2})^2 }{(\frac{\sqrt{5}}{3})^2+1} \\=\frac{\frac{4}{3}-\frac{5}{4}}{\frac{5}{9}+1} \\=\frac{\frac{16-15}{12} }{\frac{5+9}{9} } \\=\frac{3}{56}$

Therefore,

The value of the given trigonometric expression is

$\frac{2cos\theta-tan^2\theta}{sin^2\theta+1}=\frac{3}{56}$

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