Simplify sqrt(1 + tan^2(alpha) * sin^2(beta)).

Step-by-step solution

1. 1

   Resist the reflex $1+\tan^2=\sec^2$. That identity needs the tangent term to stand alone; here it is multiplied by $\sin^2\beta$, so $1+\tan^{2}\alpha\sin^{2}\beta$ is not $\sec^{2}\alpha\sin^{2}\beta$ or anything so tidy. The reliable route is to convert to sines and cosines and combine.

2. 2

   Rewrite the tangent. With $\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$,

   $1+\tan^{2}\alpha\sin^{2}\beta=1+\frac{\sin^{2}\alpha\sin^{2}\beta}{\cos^{2}\alpha}$

3. 3

   Put everything over the common denominator $\cos^{2}\alpha$.

   $=\frac{\cos^{2}\alpha+\sin^{2}\alpha\sin^{2}\beta}{\cos^{2}\alpha}$

   A single fraction is what lets the square root split cleanly in the next step.

4. 4

   Take the square root, keeping the absolute value. Since $\sqrt{\cos^{2}\alpha}=\left|\cos\alpha\right|$ (not $\cos\alpha$, which can be negative):

   $\sqrt{1+\tan^{2}\alpha\sin^{2}\beta}=\frac{\sqrt{\cos^{2}\alpha+\sin^{2}\alpha\sin^{2}\beta}}{\left|\cos\alpha\right|}$

5. 5

   Simplify the numerator with the Pythagorean identity. Replacing $\cos^{2}\alpha=1-\sin^{2}\alpha$ and factoring:

   $\cos^{2}\alpha+\sin^{2}\alpha\sin^{2}\beta=1-\sin^{2}\alpha\left(1-\sin^{2}\beta\right)=1-\sin^{2}\alpha\cos^{2}\beta$

   so an equivalent and more symmetric form is

   $\frac{\sqrt{1-\sin^{2}\alpha\cos^{2}\beta}}{\left|\cos\alpha\right|}$

6. 6

   Check with numbers. At $\alpha=0.9,\ \beta=2.1$ the original evaluates to $1.477588$ and both simplified forms give $1.477588$ ✓. The second form also makes the range obvious: the numerator lies between $\left|\cos\alpha\right|$ and $1$, so the whole expression is never below $1$.