Question

sin theta + cosec theta}whole square + cos theta + sec theta }whole square = ???

Answer 1

Conditions

{sin theta + cosec theta}whole square + {cos theta + sec theta}whole square = ???

Solution

Let's simplify it:

$$\begin{array}{l} (sin\theta + csc\theta)^2 + (cos\theta + sec\theta)^2 \\ = sin^2\theta + 2sin\theta \frac{1}{sin\theta} + \frac{1}{sin^2\theta} + cos^2\theta + 2cos\theta \frac{1}{cos\theta} + \frac{1}{cos^2\theta} \end{array}$$ $$sin^2\theta + cos^2\theta = 1$$ $$2sin\theta \frac{1}{sin\theta} = 2$$ $$2cos\theta \frac{1}{cos\theta} = 2$$

Then:

$$sin^2\theta + 2sin\theta \frac{1}{sin\theta} + \frac{1}{sin^2\theta} + cos^2\theta + 2cos\theta \frac{1}{cos\theta} + \frac{1}{cos^2\theta} = 1 + 2 + 2 + \frac{1}{sin^2\theta} + \frac{1}{cos^2\theta}$$ $$5 + \frac{1}{sin^2\theta} + \frac{1}{cos^2\theta} = 5 + sec^2\theta + csc^2\theta$$