Question

$\frac{sin\Theta}{1-cot\Theta}+\frac{cos\Theta}{1-tan\Theta}$ is equal to
(a)0
(b)1
(c)sin θ + cos θ
(d)sin θ − cos θ

Answer 1

Answer:

sinθ/(1 − cotθ) + cosθ/(1 − tanθ) is equal to sinθ + cosθ

Among the given options option (c)  sinθ + cosθ  is correct.  

Step-by-step explanation:

Given:  sinθ/(1 − cotθ) + cosθ/(1 − tanθ)

= sinθ/(1 − cosθ/sinθ) + cosθ/(1 − sinθ/cosθ)  

[By using the identity, cotθ = cosθ/sinθ  ,  tanθ = sinθ/cosθ ]  

= sinθ/[(sinθ − cosθ)/sinθ] + cosθ/[(cosθ− sinθ)/cosθ)]  

[By taking LCM]  

 

= sinθ × sinθ / [(sinθ − cosθ)] + cosθ × cosθ /[(cosθ− sinθ)]  

=sin² θ /(sinθ − cosθ) -  cos²θ/(sinθ - cosθ−)

= (sin² θ - cos²θ) / (sinθ − cosθ)  

[By taking LCM]  

= [(sinθ + cosθ) (sinθ − cosθ) ] /(sinθ − cosθ)  

[By using identity , a² - b² = (a + b) (a - b)]  

= (sinθ + cosθ)  

Hence,  sinθ/(1 − cotθ) + cosθ/(1 − tanθ) is equal to sinθ + cosθ

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Answer:

$\displaystyle{\implies\sin\theta+\cos \theta}$

Option c.  is correct .

Step-by-step explanation:

Given :

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

$\displaystyle{\implies\dfrac{\sin^2\theta}{\sin\theta-\cos \theta}-\dfrac{\cos^2\theta}{\sin\theta-\cos\theta}}}\\\\\\\displaystyle{\implies\dfrac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos \theta}}\\\\\\\displaystyle{\implies\dfrac{(\sin\theta-\cos \theta)(\sin\theta+\cos \theta)}{(\sin\theta-\cos \theta)}}\\\\\\\displaystyle{\implies\sin\theta+\cos \theta}$

Thus we get answer .