Question 175

If θ is an acute angle and $$\tan^2\theta+\frac{1}{\tan^2\theta}=2$$ then the value of θ is :

Solution

Expression : $$\tan^2\theta+\frac{1}{\tan^2\theta}=2$$

=> $$(tan\theta + \frac{1}{tan\theta})^2 - 2 = 2$$

=> $$(tan\theta + \frac{1}{tan\theta})^2 = 4$$

=> $$tan\theta + \frac{1}{tan\theta} = 2$$

[It can't be -2 as $$\theta$$ is in 1st quadrant, and $$tan\theta$$ is positive in 1st quadrant.]

=> $$tan^2\theta + 1 = 2tan\theta$$

=> $$(tan\theta - 1)^2 = 0$$

=> $$tan\theta = 1$$

=> $$\theta = 45^{\circ}$$

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