CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$\displaystyle 0^\circ<\theta < 90^{\circ}$$ and $$\displaystyle \frac{\sin \theta }{1-\cos \theta }+\frac{\sin \theta }{1+\cos \theta }=4,$$ then the value of $$\displaystyle \theta $$ is 


A
30
loader
B
45
loader
C
60
loader
D
none
loader

Solution

The correct option is A 30$$\displaystyle ^{\circ}$$

Given, $$ \dfrac {sin \theta}{1-cos \theta} + \dfrac {sin \theta}{1+cos \theta} =4 $$
Taking LCM and cross multiplying we get 
$$ \dfrac {sin \theta(1+cos \theta + 1 - cos \theta)}{(1-cos \theta)(1+cos \theta)} = 4 $$
$$ => \dfrac {sin \theta(1+1)}{(1-cos^2 \theta)} = 4 $$
$$ => \dfrac {2sin \theta)}{(sin^2 \theta)} = 4 $$   (Since $$ 1-cos^2 \theta = sin^2 \theta ) $$
$$ => sin \theta  = \dfrac {1}{2} $$
$$ => \theta = 30^0 $$


Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image