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
B
45
C
60
D
none

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

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