Question

If 3 cot A = 4, check whether (1 – tan2A)/(1 + tan2A) = cos2 A – sin2 A or not.​

Answer 1

Answer:

f 3 cot A = 4, check whether (1 – tan2A)/(1 + tan2A) = cos2 A – sin2 A or not.​

Step-by-step explanation:

Answer 2

Given:

• 3 cot A = 4 => cot A = 4/3

Solution

Let us consider a triangle ABC, im which right-angled at B

Since,

tan A = 1/cot A

tan A = 1 /(4/3) = 3/4

_____________

${\tt{ \dfrac{BC}{AB} = \dfrac{3}{4} }}$

Let BC = 3k and AB = 4k

By using Pythagoras theorem,

(Hypotenuse)² = (Perpendicular)² + (Base)²

$\implies$ AC² = AB² + BC²

$\implies$ AC² = (4k)² + (3k)²

$\implies$ AC² = 16k² + 9k²

$\implies$ AC = √25k²

$\implies$AC = 5k

$\\ \bigstar{\pink{\boxed {\tt{ sin \ A = \dfrac{Opposite \ side}{Hypotenuse} }}}}$

$\implies$ BC/AC

$\implies$ 3k/5k

$\implies$ 3/5

Same as the above;

$\\ {\tt{ cos \ A = \dfrac{Adjacent \ side}{Hypotenuse} }}$

$\implies$ AB/AC

$\implies$ 4k/5k

$\implies$ 4/5

______________

To check:

• (1-tan²A)/(1+tan²A) = cos² A – sin² A or not

Let take L.H.S. first;

$\implies$ (1-tan²A)/(1+tan²A)

$\implies$ [1 – (3/4)²]/ [1 + (3/4)²]

$\implies$ [1 – (9/16)]/[1 + (9/16)]

$\implies$ 7/25

Now, take R.H.S. ;

$\implies$ cos² A – sin² A

$\implies$ (4/5)² – (3/5)²

$\implies$ (16/25) – (9/25)

$\implies$ 7/25

Since,

L.H.S. = R.H.S.

${\red{\boxed{\underline {\tt{ \dfrac{7}{25} _{(LHS)} = \dfrac{7}{25} _{(RHS)} }}}}}$

Hence proved ..!