Question

Pls do it... only no.3

Answer 1

Answer:

sin²θ + cosec²θ = 1

Step-by-step explanation:

 ⇒ sinθ + cosecθ = √3

Square on both sides :

⇒ ( sinθ + cosecθ )^2 = ( √3 )^2

⇒ sin²θ + cosec²θ + 2( sinθ * cosecθ ) = 3

  sinθ*cosecθ = sinθ*(1/sinθ ) = 1

⇒ sin²θ + cosec²θ + 2(1) = 3

⇒ sin²θ + cosec²θ = 3 - 2

⇒ sin²θ + cosec²θ = 1

    Hence proved.

Answer 2

Given :

$\sf \star \: \: Sin( \theta) + Cosec( \theta) = \sqrt{3}$

To prove :

$\sf \star \: \: {Sin}^{2} ( \theta) + {Cosec}^{2} ( \theta) = 1$

Proof :

LHS :

$\Rightarrow \sf Sin( \theta) + Cosec( \theta) = \sqrt{3} \\ \\ \: \: \: \: \sf Squaring \: both \: sides \: , \: we \: get \\ \\ \Rightarrow \sf {(Sin ( \theta) + Cosec( \theta))}^{2} = { (\sqrt{3} )}^{2} \\ \\ \Rightarrow \sf {Sin}^{2} ( \theta) + {Cosec}^{2} ( \theta) + 2(Sin ( \theta) \times Cosec( \theta) ) = 3 \\ \\ \Rightarrow \sf {Sin}^{2} ( \theta) + {Cosec}^{2} ( \theta) + 2(1) = 3 \: \: \{ \because Sin ( \theta) \times Cosec( \theta) = 1\} \\ \\ \Rightarrow \sf {Sin}^{2} ( \theta) + {Cosec}^{2} ( \theta) = 1$

$\large{ \star} \: \: \sf LHS = RHS$

$\sf \underline{ \bold{ \: Hence \: proved \: }}$