Question

cot theta =2x/root 9-4x^2 evaluate. sin theta +cos theta and cosec theta-tan theta

Answer 1

there you go young boy here's your answer

Answer 2

sin theta +cos theta=2 x + root9 − 4 x^2/3

cosec theta-tan theta=6 x + 9 − 4 x^2/2 x root 9 − 4 x^2

Step-by-step explanation:

Given:

$cot \theta = \frac{2x}\sqrt{9-4x^2}}$

We know that,

$cot\theta=\frac{Adjacent}{Opposite}$

We know from Pythagoras theorem:

$Hypotenuse^2 = Adjacent^2 + Opposite^2$

$Hypotenuse^2=(2x)^{2}+(\sqrt{9 - 4 x^2} )^2$

$Hypotenuse^2 = 4 x^2 + 9 - 4 x^2$

$Hypotenuse^2 = 9$

$Hypotenuse = \sqrt{9}$

Hypotenuse =3

And we know,

$sin\theta=\frac{Opposite}{Hypotenuse}$

So, $sin\theta=\frac{\sqrt{9-4x^2}}{3} }$

$cos\theta=\frac{Adjacent}{Hypotenuse}$

So, $cos\theta=\frac{2x}{3}$

$cosec\theta=\frac{Hypotenuse}{Opposite}$

So, $cosec\theta=\frac{3}{\sqrt{9-4x^} }$

$tan\theta=\frac{Opposite}{Adjacent}$

So,$tan\theta=\frac{\sqrt{9-4x^2} }{3}$

So,

1) We find value of $sin \theta+ cos \theta$ by substituting values, As :

$\frac{\sqrt{9-4x^2} }{3} +\frac{2x}{3}$

=$\frac{2x+\sqrt{9-4x^2} }{3}$

2) We find value of $cosec\theta + tan\theta$ by substituting values, As :

$\frac{3}{\sqrt{9-4x^2} } +\frac{\sqrt{9-4x^2} }{2x}$

=$\frac{3\times2x+(\sqrt{9-4x^2)^2} }{2x\sqrt{\sqrt{9-4x^2} } }$

=$\frac{6 x + 9 - 4 x^2}{2x\sqrt{9-4x^2} }$