Question

Answer those two questions....​

Answer 1

1] Given : 3cosθ = 5sinθ ...(i)

To Find : $\dfrac{5sinθ - 2 {sec}^{3} θ + 2cosθ}{5sinθ + 2 {sec}^{3} θ - 2cosθ}$ ...(ii)

Now,

sinθ/cosθ = 3/5 [using(i)]

tanθ = 3/5

We know that, tanθ = Perpendicular/Base

So, Let perpendicular = 3x, Base = 5x

By Pythagoras Theorem,

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

Hypotenuse = √(3x)² + (5x)²

= √9x² + 25x²

= √34x²

= √34

sinθ = Perpendicular/Hypotenuse = 3/√34

cosθ = Base/Hypotenuse = 5/√34

secθ = 1/cosθ = √34/5

Now, putting the known values in (ii), we get

→ $\dfrac{5( \dfrac{3}{ \sqrt{34} }) - 2 {( \dfrac{ \sqrt{34} }{5} )}^{3} + 2( \dfrac{5}{ \sqrt{34} } ) }{5( \dfrac{3}{ \sqrt{34} }) + 2 {( \dfrac{ \sqrt{34} }{5} )}^{3} - 2( \dfrac{5}{ \sqrt{34} } )}$

→ $\dfrac{ \dfrac{15}{ \sqrt{34} } - \dfrac{68 \sqrt{34} }{125} + \dfrac{10}{ \sqrt{34} } }{ \dfrac{15}{ \sqrt{34} } + \dfrac{68 \sqrt{34} }{125} - \dfrac{10}{ \sqrt{34} }}$

→ $\dfrac{ \dfrac{15(125) - 68 \sqrt{34} ( \sqrt{34}) + 10(125) }{( \sqrt{34} )(125)} }{ \dfrac{15(125) + 68 \sqrt{34}( \sqrt{34)} - 10(125)}{ \sqrt{34} )(125)} }$

→ $\dfrac{ \dfrac{1875 - 2312 + 1250}{125 \sqrt{34} } }{ \dfrac{1875 + 2312 - 1250}{125 \sqrt{34} } }$

→ $\dfrac{813}{2937}$

→ 271/979

2] Given : 3sinx + 5cosx = 5

To Find : (3cosx - 5sinx)²

Now,

As we know that,

sin²x + cos²x = 1 ...(i)

3sinx + 5cosx = 5 (given)

Squaring both sides, we get

→ (3sinx + 5cosx)² = 5²

→ 9sin²x + 25cos²x + 30sinx.cosx = 25

...(ii)

Let, (3cosx - 5sinx)² = p

On solving further, we get

→ 9cos²x + 25sin²x - 30sinx.cosx = p

...(iii)

Adding (i) & (ii), we get

9(sin²x + cos²x) + 25(sin²x + cos²x) = p + 25

9(1) + 25(1) = p + 25 [using(1)]

9 + 25 = p + 25

p = 9

So, (3cosx - 5sinx)² = 9