Question

If secA = 5/4 then prove that tanA/1+tan^(2)A = sinA/secA​

Answer 1

Step-by-step explanation:

secA=AC/AB=5/4

In above figure, ∆ABC and angle B=90°.

By Pythagoras theorem:

AC²=AB²+BC²

5²-4²=BC²

25-16=BC²

√9=3=BC.

tanA=BC/AB=3/4.

tan²A=BC²/AC²=3²/4²=9/16.

cosA=AB/AC=4/5

sinA=BC/AC=3/5

LHS=tanA/1+tan²A

$= \frac{3}{ \frac{4}{1 + \frac{9}{16} } } = > \frac{3}{ \frac{4}{ \frac{16 + 9}{16} } }$

$= \frac{3 \times 16}{4(16 + 9)} = > \frac{3 \times 4}{25} = > \frac{12}{25}$

RHS=sinA/secA

$= \frac{3}{ \frac{5}{ \frac{5}{4} } } = > \frac{3 \times 4}{5 \times 5} = > \frac{12}{25}$

LHS=RHS.

Hope helps u.