Question

Find x from the following equations :

i -> Cosec ( 90 + Φ ) + x Cos Φ Cot ( 90 + Φ ) = Sin ( 90 + Φ )

ii -> x Cot ( 90 + Φ ) + Tan ( 90 + Φ ) Sin Φ + Cosec ( 90 + Φ ) = 0​

Answer 1

Answer:

$\displaystyle{1). \ x=\tan\theta}\\\\\\\displaystyle{2). \ x=\sin\theta}$

Step-by-step explanation:

Given :

Cosec ( 90 + Φ ) + x Cos Φ Cot ( 90 + Φ ) = Sin ( 90 + Φ )

sec  Φ + x cos  Φ - tan  Φ = cos  Φ

-  x cos  Φ tan  Φ =  cos  Φ - sec Φ

x = cos  Φ - sec Φ / - cos  Φ tan  Φ

x = sec Φ - cos Φ /  cos  Φ tan  Φ

$\displaystyle{\implies\dfrac{\dfrac{1}{\cos \theta}- \cos \theta}{\cos\theta\times\dfrac{\sin\theta}{\cos \theta}}}\\\\\\\displaystyle{\implies\dfrac{\dfrac{1}{\cos \theta}- \cos \theta}{\sin\theta}}\\\\\\\displaystyle{\implies\dfrac{\dfrac{1-\cos^2 \theta}{\cos \theta}}{\dfrac{\sin\theta}{1}}}\\\\\\\displaystyle{\implies\dfrac{\sin^2\theta}{\sin\theta\times\cos\theta}}\\\\\\\displaystyle{\implies\dfrac{\sin\theta}{\cos\theta}}\\\\\\\displaystyle{\implies\tan\theta}$

x = tan  Φ

ii .

x Cot ( 90 + Φ ) + Tan ( 90 + Φ ) Sin Φ + Cosec ( 90 + Φ ) = 0​

- x  tan  Φ  -  cot  Φ  sin  Φ + sec  Φ = 0

-  cot  Φ  sin  Φ + sec  Φ =  x  tan  Φ

x =  sec  Φ  -  cot  Φ  sin  Φ /  tan  Φ

$\displaystyle{\implies\dfrac{\dfrac{1}{\cos \theta}- \cos \theta}{\dfrac{\sin\theta}{\cos \theta}}}\\\\\\\displaystyle{\implies\dfrac{\dfrac{1-\cos^2\theta}{\cos \theta}}{\dfrac{\sin\theta}{\cos \theta}}}\\\\\\\displaystyle{\implies\dfrac{1-\cos^2\theta\times \cos\theta}{\sin\theta\times\cos\theta}}\\\\\\\displaystyle{\implies\dfrac{\sin^2\theta}{\sin\theta}}\\\\\\\displaystyle{\implies \sin\theta}$

x = sin Φ

Thus we get answer .

Answer 2

hey sagar...

here is your answer..

refer to the aatachment