Question

If tan A = 24, find the value of sin A + cos A​

Answer 1

Answer:

sinA+cosA

tanA=sinA/cosA

as, tanA=24

so, sinA/cosA=24

sin=cos24

cos=sin/24

substitute, sinA+cosA

cos24+sin/24=0

24/24(cos+sin)=0

cos+sin=1

1*1=1

sinA+cosA=1

ans=1

Answer 2

GIVEN :–

$\\ \bf \implies \tan(A) = 24$

TO FIND :–

• Value  of  sin(A) + cos(A) = ?

SOLUTION :–

$\\ \bf \implies \tan(A) = 24$

• We  know  that  –

$\\ \bf \implies \tan(A) = \dfrac{L}{A}$

$\\ \bf \implies \tan(A) = \dfrac{L}{A} =  \dfrac{24}{1}$

• So that  –

$\\ \bf \implies L = 24 \:, \:  A = 1$

• We also that –

$\\ \bf \implies  {K}^{2} = L^{2} + A ^{2}$

$\\ \bf \implies  {K}^{2} = (24)^{2} + (1)^{2}$

$\\ \bf \implies K =  \sqrt{576+1}$

$\\ \bf \implies K =  \sqrt{577}$

• We also know that  –

$\\ \bf \implies \sin(A) = \dfrac{L}{K}$

• And –

$\\ \bf \implies \cos(A) = \dfrac{A}{K}$

• Now let's find –

$\\ \bf \:   \: =  \:  \: \sin(A)+ \cos(A)$

$\\ \bf \:   \: =  \:  \:\dfrac{L}{K}+ \dfrac{A}{K}$

$\\ \bf \:   \: =  \:  \:\dfrac{L+A}{K}$

• Now put the values –

$\\ \bf \:   \: =  \:  \:\dfrac{24+1}{\sqrt{577}}$

$\\ \bf \:   \: =  \:  \:\dfrac{25}{\sqrt{577}}$

• Hence –

$\\ \large \implies{ \boxed{ \bf\sin(A) + \cos(A)=\dfrac{25}{\sqrt{577}}}}$