Question

If sin theta= 4/5, find the value of (sin theta tan theta-1)/2tan^2theta

Answer 1

Step-by-step explanation:

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See The Attachment My Friend....

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Answer 2

Answer:

The value of the given trigonometric relation is

$\frac{sin\theta tan\theta-1}{2tan^{2}\theta} = \frac{3}{160}$

Step-by-step explanation:

The given trigonometric relation is

$sin\theta = \frac{4}{5}$

We are required to find the other trigonometric relation for which we are required to find the tangent ratio of the angle based of the sine ratio.It is found in the way as shown below

$sin \theta = \frac{4}{5} \\ = > cos \theta = \sqrt{1 - {sin}^{2} \theta } \\ = > cos \theta = \sqrt{1 - ( \frac{4}{5} ) {}^{2} } = \frac{3}{5}$

The tan ratio is the ratio of sine to cosine of the angle.It is calculated as shown

$tan \theta = \frac{sin \theta}{cos \theta} = \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{3}$

The required trigonometric relation to be calculated is

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

Substituting the known trigonometric ratios in above equation,we get the relation as

$\frac{ \frac{4}{5} \times \frac{4}{3} - 1}{2( \frac{4}{3} ) {}^{2} } = \frac{ \frac{16}{15} - 1 }{2 \times \frac{16}{9} } \\ = \frac{ \frac{1}{15} }{ \frac{32}{9} } = \frac{9}{15 \times 32} = \frac{3}{160}$

Therefore,the value of given trigonometric relation is

$\frac{sin\theta tan\theta-1}{2tan^{2}\theta} = \frac{3}{160}$

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