Question

if sin theta is equal to 4 by 5 find the value of sin theta tan theta minus 1 upon 2 tan square theta...... please its urgent

Answer 1

look at atttachment ………

ans
$\frac{3}{160}$

Answer 2

Answer:

The value of $\frac{sin\theta tan\theta - 1}{2tan^{2}\theta }$ is $\frac{3}{160}$ .

Step-by-step explanation:

Explanation:

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

and  also given that $\frac{sin\theta tan\theta - 1}{2tan^{2}\theta }$

As we know , $sin\theta = \frac{p}{h}$

and $tan \theta = \frac{p}{b}$

Step 1:

We have  sin$\theta$ = $\frac{4}{5}$  ,

 Here perpendicular (p) = 4 and hypotenuse (h) = 5

∴ Base (b) = $\sqrt{h^{2} - p^{2} }$ = $\sqrt{5^{2} - 4^{2} } = \sqrt{25 - 16}$

⇒b= $\sqrt{9} = 3$

$tan \theta = \frac{p}{b} = \frac{4}{3}$

Step 2:

Now , we have from the question

$\frac{sin\theta tan\theta - 1}{2tan^{2}\theta }$   ..........(i)

Now put the value of $sin\theta \ and tan \theta$ in (i) we get ,

⇒$\frac{sin\theta tan\theta - 1}{2tan^{2}\theta }$ = $\frac{\frac{4}{5} \frac{4}{3} -1}{2(\frac{4}{3} )^{2} }$ = $\frac{\frac{16}{15} - 1 }{\frac{32}{9} }$

⇒$\frac{sin\theta tan\theta - 1}{2tan^{2}\theta }$ = $\frac{\frac{1}{15} }{\frac{32}{9} }$ = $\frac{9}{ 15 . 32}$ = $\frac{3}{160}$ .

Final answer:

Hence , the value of $\frac{sin\theta tan\theta - 1}{2tan^{2}\theta }$ is $\frac{3}{160}$ .

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