Question 56

$$\left(\frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \cosec^2 \theta}\right)$$ =

Solution

$$\left(\frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \cosec^2 \theta}\right)$$

where Cosec(t) = $$\frac{1}{sint}$$ transform the expression

=$$\frac{1}{1+sin^2t}$$ + $$\frac{1}{1+ (1÷sin^2t)}$$

=$$\frac{1}{1+sin^2t}$$ + $$\frac{1}{(sin^2t+1÷sin^2t)}$$

=$$\frac{1}{1+sin^2t}$$ + $$\frac{sin^2t}{sin^2t+1}$$ take the orders of this then reduce simplification is

=$$\frac{1+ sin^2t}{1+sin^2t}$$

= 1

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