Question

in right angled triangle angle A = o and angle B = 90° 3 tan =4 verify that​

Answer 1

Step-by-step explanation:

We have

3 tan ø = 4 ,

⇒tan ø=4/3

Now tan ø=3/4= perpendicular/base =p/b

Let p=3x and b=4x ,

Also let h = hypotenuse

• By pythagorus theorum

h²=p²+b²

h²= (3x)²+(4x)²

h²= 9x²+16x²

h²= 25 x²

h = 5x

Thus tan ø= p/b = 3/4

and sin ø= h/b =3/5

We have LHS

$\sf \implies \dfrac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta }$

$\sf \implies \dfrac{1 - \dfrac{3^{2} }{4^{2} } }{1 + \dfrac{3^{2} }{4^{2} } }$

$\sf \implies \dfrac{1 - \dfrac{9}{16 } }{1 + \dfrac{9 }{16} }$

$\sf \implies \dfrac{\dfrac{16 - 9}{ \cancel{16} } }{ \dfrac{16 + 9 }{ \cancel{16} } }$

$\sf \implies \: \dfrac{7}{25}$

We have RHS

$\sf \implies1 - 2 sin^{2} \theta$

$\sf \implies1 - 2 \times \dfrac{3^{2} }{5^{2} }$

$\sf \implies1 - 2 \times \dfrac{9 }{25}$

$\sf \implies1 - \dfrac{ 18}{25 }$

$\sf \implies\dfrac{ 25 - 18}{25 }$

$\implies \sf \dfrac{7}{25}$

LHS=RHS

Hence, verified