Question

Choose the correct option of general solutions -
$$\begin{array}{cc}TrigonometricEquations& GeneralSolutions\\ 1.si{n}^2\theta =si{n}^{2}x& A.2n\pi \pm \alpha \\ 2.co{s}^2\theta =co{s}^{2}x& B.n\pi +(-1{)}^n\alpha \\ 3.ta{n}^2\theta =ta{n}^{2}x& C.n\pi \pm \alpha \\ & \end{array}$$

• A. 1 - A, 2 - B, 3 - C
• B. 1 - B, 2 - A, 3 - C
• C. 1 - A, 2 - A, 3 - A
• D. 1 - C, 2 - C, 3 - C

Answer 1

The correct option is D

1 - C, 2 - C, 3 - C

1 . $si{n}^2$θ = $si{n}^2$α

$si{n}^2$θ - $si{n}^2$α = 0

= sin (θ+α) × sin (θ-α) = 0

So, sin (θ+α) = 0 or sin (θ-α) = 0

θ + α = nπ or θ - α = nπ

θ = nπ - α or θ = nπ + α

Combining both the equations,

We get, θ = nπ ± α

2. $co{s}^2$θ = $co{s}^2$α

$co{s}^2$θ - $co{s}^2$α = 0

- sin (θ+α) sin (θ-α) = 0

sin (θ+α) =0 or sin (θ-α) = 0

θ + α = nπ or θ - α = nπ

θ = nπ - α or θ = nπ + α

Combining both the equations,

We get, θ = nπ ± α

3. $ta{n}^2$θ = $ta{n}^2$α

$\frac{si{n}^2\theta }{co{s}^2\theta }$ - $\frac{si{n}^2\alpha }{co{s}^2\alpha }$ = 0

$\frac{si{n}^2\theta .co{s}^2\alpha -co{s}^2\theta .si{n}^2\alpha }{co{s}^2\theta .co{s}^2\alpha }$ = 0

$\frac{(sin\theta .cos\alpha {)}^2-(cos\theta .sin\alpha {)}^2}{co{s}^2\theta .co{s}^2\alpha }$ = 0

$\frac{(sin\theta .cos\alpha +cos\theta .sin\alpha )(sin\theta .cos\alpha -cos\theta .sin\alpha )}{co{s}^2\theta .co{s}^2\alpha }$ = 0

sin(θ +α) sin(θ -α) = 0 { where θ&α should not be equal to (2n+1)$\frac{\pi }{2}$ n ∈ Z}

sin (θ+α) =0 or sin (θ-α) = 0

θ + α = nπ or θ - α = nπ

θ = nπ - α or θ = nπ + α

Combining both the equations,

We get, θ = nπ ± α

Where n ∈ Z, α & θ should not be equal odd multiples of π /2

α & θ ≠ (2n+1) π /2 ∈ Z