Question

Please answer question 80

Answer 1

Step-by-step explanation:

Here 4cosθ - 3sinθ = 5

Method 1. Using Differentiation

Differentiating the above equation,we get

4* d(cosθ)/dθ - 3* d(sinθ)/dθ = d(5)/dθ

⇒ -4sinθ - 3cosθ = 0

⇒ 4sinθ + 3cosθ = 0

Hence proved.

Method 2. Using tigonometry

Trigonometric Identity used

1. cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

2. sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

Now 4cosθ - 3sinθ = 5

⇒ cosθ*(4/5) - 3sinθ*3/5 = 1

⇒ cosθ*cosφ- sinθ*sinφ= 1

where cosφ = 4/5, sinφ = 3/5

⇒ cos(θ+φ) = 1 = cos(0)

⇒ θ +φ = 0

And LHS = 4sinθ + 3cosθ

= 5*(sinθ*(4/5) + cosθ*(3/5))

= 5*(sinθ*cosφ + cosθ*sinφ)

= 5*sin(θ+φ)

= 5*sin(0)

= 5*0

= 0 = RHS

Hence Proved.

Answer 2

Answer:

$\begin{gathered}\begin{gathered}\\\\\sf \large \red{\underline{Question:-}}\\\\\end{gathered}\end{gathered}</p><p>$

If 3sinθ+4cosθ=5,then prove that

4sinθ−3cosθ=0

$\begin{gathered}\begin{gathered}\\\\\sf \large \red{\underline{Solution :- }}\\\\\end{gathered}\end{gathered}</p><p>$

We have,

3sinθ + 4cosθ = 5 ------------------1

On squaring on both sides,

( 3sinθ + 4cosθ )² = 5²

9sin²θ + 16cos²θ + 24sinθcosθ = 25

9(1−cos²θ)+16(1−sin²θ)+12×2sinθcosθ=25

9−9cos²θ+16−16sin²θ+12×2sinθcosθ=25

25−9cos²θ−16sin²θ+12×2sinθcosθ=25

9cos²θ + 16sin²θ − 12×2sinθcosθ = 0

(3cosθ−4sinθ)²=0

3cosθ−4sinθ=0

$\tt \red{hence \: it \: is \: proved}$