$$\sec^2 29^\circ - \cot^2 61^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$ is equal to:
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, ( $$\cosec^{2}\theta-\cot^{2}\theta=1$$)
  = $$\frac{23}{4}$$
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