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

1 + $\frac{t a n A}{t a n B}$ =

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$s e c^{2} A$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

secA

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$s e c^{2} A$

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

So, 1 + $\frac{t a n (90 - B)}{t a n B}$

= 1 + $\frac{c o t B}{t a n B}$
We know $\frac{1}{t a n B} = c o t B$

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

= 1 + $c o t^{2} (90 - A)$

= 1 + $t a n^{2} A$

= $s e c^{2} A$ ( $s e c^{2} A - t a n^{2} A = 1$ )

Suggest Corrections

Similar questions

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

1 + $\frac{t a n A}{t a n B}$ =

\sum \frac{c o t A + c o t B}{t a n A + t a n B} =

△ A B C

A

B

tan A + tan B + {tan}^{2} A + {tan}^{2} B = 10

△ A B C

tan A + tan B + tan C

Join BYJU'S Learning Program