Question 69

$$\sec^2 29^\circ - \cot^2 61^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$ is equal to:

Solution

We know that $$\sec(90^\circ-\theta)$$ = $$\cosec\theta$$

$$\sec29^\circ=\cosec61^\circ$$

Now, $$\cosec^{2}61^\circ-\cot^{2}61^\circ+\sin^{2}60^\circ+\cosec^{2}30^\circ$$

= $$1+\frac{3}{4}+\frac{4}{1}$$ Since, ( $$\cosec^{2}\theta-\cot^{2}\theta=1$$)

= $$\frac{23}{4}$$

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: