Question

If in a right-angled triangle ABC angles A and B are acute, then evaluate

1 + $\frac{tanA}{tanB}$ =

• A. 1
• B. $se{c}^{2}A$
• C. secA
• D. 2

Answer 1

The correct option is B

$se{c}^{2}A$

Given, in a right-angled triangle ABC angles A and B are acute.
Hence, A + B = ${90}^{\circ }$.

So, 1 + $\frac{tan(90°-B)}{tanB}...\left[\tan \right(90-\theta )=\cot \theta$]

= 1 + $\frac{cotB}{tanB}$
We know $\frac{1}{tanB}=cotB$

= 1 + ${\cot }^{2}B$

= 1 + $co{t}^2(90°-A)$

= 1 + $ta{n}^{2}A$

= $se{c}^{2}A$ ($se{c}^{2}A-ta{n}^{2}A=1$)