Question

if tanA=5/2 then prove 2cos2A+ 5sin2A=2

Answer 1

Answer:MARK ME BRAINLIST

Heya user ,

Here is your answer !!

5 sin²A + 2 cos²A = tanA=5/2

=> 5 sin²A + 2 cos²A + 4 sin²A = tanA=5/2

=> 5 ( sin²A + cos²A ) + 4 sin²A =tanA=5/2

=> 5 + 4 sin²A = tanA=5/2

=> 5 sin²A = 1

=> sin²A =tanA=5/2

=> sin A = 1/2

=> sin A = tanA=5/2

=> A = tanA=5/2

then  2cos2A+ 5sin2A=2

Hope it helps !!

Step-by-step explanation:

Answer 2

Answer:

$\star\:\sf \tan(A)=\dfrac{5}{2}=\dfrac{Perpendicular}{Base}$

$\dashrightarrow\sf\:\:(Hypotenuse)^2=(Perpendicular)^2+(Base)^2\\\\\\\dashrightarrow\sf\:\:(h)^2=(p)^2+(b)^2\\\\\\\dashrightarrow\sf\:\:(h)^2=(5)^2+(2)^2\\\\\\\dashrightarrow\sf\:\:(h)^2 = 25 + 4\\\\\\\dashrightarrow\sf\:\:(h)^2 = 29\\\\\\\dashrightarrow\sf\:\:h = \sqrt{29}$

$\rule{160}{2}$

$\underline{\bigstar\:\boldsymbol{According\:to\:the\:Question :}}$

$:\implies\sf 2\cos(2A)+5\sin(2A)=2\\\\\\:\implies\sf 2[\cos^2(A) - \sin^2(A)]+5[2 \sin(A) \cos(A)] = 2\\\\\\:\implies\sf 2\bigg\lgroup\left(\dfrac{2}{ \sqrt{29} }\right)^2 - \left(\dfrac{5}{ \sqrt{29} }\right)^2\bigg\rgroup+5\bigg\lgroup2 \times \dfrac{5}{\sqrt{29}} \times \dfrac{2}{\sqrt{29}}\bigg\rgroup = 2\\\\\\:\implies\sf 2\left(\dfrac{4}{29} - \dfrac{25}{29} \right) + \left(5 \times\dfrac{20}{29}\right) = 2\\\\\\:\implies\sf \left(2 \times \dfrac{ -21}{29}\right) + \dfrac{100}{29} = 2\\\\\\:\implies\sf \dfrac{ -\:42}{29} + \dfrac{100}{29} = 2\\\\\\:\implies\sf \dfrac{58}{29} = 2\\\\\\:\implies\large\underline{\boxed{\sf 2 = 2}}$

${\qquad\underline{\mathcal{HENCE, PROVED}}}$