Question

Please solve ques 38 in the attachment

Answer 1

$1 + sin^{2} \theta = 3 \: sin \theta \: cos \theta$




$\text{ \underline{Divide by \: cos}}^{2} \underline{\: \theta \text{ on both sides ,}}$




$\textcolor{lightblue}{ \Rightarrow} \frac{1 + sin ^{2} \theta}{ {cos}^{2} \theta} = \frac{3 \: sin \theta \: cos \theta}{ {cos}^{2} \theta} \\ \\ \\ \\ \textcolor{blue}{ \Rightarrow} {sec}^{2} \theta + {tan}^{2} \theta = 3 \: tan \theta \\ \\ \\ \\ \\ \textcolor{darkblue}{ \Rightarrow} 1 + {tan}^{2} \theta \: + {tan}^{2} \theta \: - 3 \: tan \theta = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: | \: \: \bold{ {sec}^{2} \theta= 1 + {tan}^{2} \theta} \\ \\ \\ \\ \text{let \: tan} \theta = x \:\:\:\:\:\:\:\: [ tan^{2} \theta = x^2]$





▶ 1 + x² + x² - 3x = 0


⏩ 1 + 2x² - 3x = 0


⏩ 2x² - 3x + 1 = 0


↪ 2x² - 2x - x + 1 = 0


↪ 2x( x - 1 ) - ( x - 1 ) = 0


↪ ( x - 1 ) ( 2x - 1 ) = 0


➡ x = 1 or 2x = 1


x = 1 or x = 1 / 2






Hence,

$tan \theta \: = x = 1 \: \: \: \: \: or \: \: \: \: \: \: tan \theta \: = x = \frac{1}{2}$

Proved.

Answer 2

(38).

Given 1 + sin^2Ф = 3sinФcosФ

It can be written as:

= > 1 + sin^2Ф - 3sinФcosФ = 0

= > sin^2Ф + cos^2Ф + sin^2Ф - 2sinФcosФ - sinФcosФ = 0

= > (sin^2Ф - 2sinФcosФ + cos^2Ф) + (sin^2Ф - sinФcosФ) = 0

= > (sinФ - cosФ)^2 + sinФ(sinФ - cosФ) = 0

= > (sinФ - cosФ)[sinФ - cosФ + sinФ] = 0

= > (sinФ - cosФ)[2sinФ - cosФ] = 0

= > (sinФ - cosФ)(2sinФ - cosФ) = 0

(i)

sinФ - cosФ = 0

sinФ = cosФ

sinФ/cosФ = 1

TanФ = 1

(ii)

2sinФ - cosФ = 0

2sinФ = cosФ

sinФ/cosФ = (1/2)

TanФ = (1/2).

Hope it helps!