Question

tan^2a +cot^2a=2,ais an acute angle, then tan^3a+cot^2a=​

Answer 1

Answer:

2

Step-by-step explanation:

Given :

$Tan^2A + Cot^2A = 2$

Then, find the value of $Tan^3A + Cot^2A$

Solution :

We know that,

0° < A < 90°

&

$CotA=\frac{1}{TanA}$

Hence,

⇒ $Tan^2A + Cot^2A = 2$

⇒ $Tan^2A + \frac{1}{Tan^2A} = 2$

⇒ $\frac{Tan^4A+1}{Tan^2A} = 2$

⇒ $Tan^4A + 1= 2Tan^2A$

⇒ $Tan^4A - 2Tan^2A + 1 = 0$

⇒ $(Tan^2A)^2 - 2Tan^2A(1) + (1)^2 = 0$ Which is of the form,

(a + b)² = a² + 2ab + b²

Hence,

⇒ $(Tan^2A - 1)^2 = 0$

⇒ $Tan^2A - 1= 0$

⇒ $Tan^2A = 1$

⇒ $TanA = \sqrt{1} = 1$

⇒ A = 45°

So,

$Tan^3A + Cot^2A = (TanA)^3 + (CotA)^2 = (Tan45°)^3 + Cot45°)^2 = (1)^3 + (1)^2 = 1 + 1 = 2$