Question

Solve this problem.​

Answer 1

Correct question:

If sinA = 2/5 then find the value of 5 + 4cot²A

Solution :-

Since sinA = 2/5,

$\sf{ \dfrac{opposite \: side}{hypotenuse} = \dfrac{2}{5} }$

So,

• Opposite side = 2 units
• Hypotenuse = 5 units

By Pythagoras theorem;

BC² = AC² - AB²

→ BC² = 5² - 2²

→ BC² = 25 - 4

→ BC = √21

$\sf{cota = \dfrac{bc}{ab} }$

→ cot A = √21 ÷ 2

Substitute the value of cot A in the given question

5 + 4cot²A

$= 5 + 4 \bigg( \dfrac{ \sqrt{21}}{2} \times \dfrac{ \sqrt{21}}{2} \bigg)$

$= 5 + 4 \bigg( \dfrac{21}{4} \bigg)$

= 5 + 21

= 26

∴ 5 + 4cot²A = 26