Question

hello dear!!! plz... help me in this question

Answer 1

Answer:

In the attachment.. Hope it's helpful..

Answer 2

Answer:

Step-by-step explanation:

Given,

⇒ cosecθ - sinθ = l        ...( $\it{i}$ )

⇒ secθ - cosθ = m        ...( $\it{ii}$)

Multiply ( i ) and ( ii ),

= > ( cosecθ - sinθ )( secθ - cosθ ) = l m

$\implies \bigg\{ \dfrac{1}{\sin\theta}-\sin\theta\bigg\}\times \bigg\{\dfrac{1}{\cos\theta}-\cos\theta\bigg\}=lm\\\\\\\implies \dfrac{1-\sin^2 \theta}{\sin\theta}\times \dfrac{1-\cos^2\theta}{\cos\theta}=lm\\\\\\\implies \dfrac{\cos^2\theta}{\sin\theta}\times \dfrac{\sin^2 \theta}{\cos\theta}=lm\quad\quad\bigg|\mathsf{1-sin^2A =cos^2A\:\:\:And\:\:\:1-cos^2A = sin^2A}\\\\\\\implies \sin\theta .\cos\theta=lm$

Square on both sides,

= > sin^2 θ . cos^2 θ = l^2 m^2    ...( iii )

Now, square on both sides of ( i ),

= > ( cosecθ - sinθ )^2 = l^2

= > cosec^2 θ + sin^2 θ - 2cosecθsinθ = l^2

= > cosec^2 θ + sin^2 θ - 2( cosecθ x 1 / cosecθ ) = l^2

= > cosec^2 θ + sin^2 θ - 2 = l^2     ...( iv )

Square on both sides of ( ii ),

= > ( secθ - cosθ )^2 = m^2

= > sec^2 θ + cos^2 θ - 2secθcosθ = m^2

= > sec^2 θ + cos^2 θ - 2 ( secθ x 1 / secθ ) = m^2

= > sec^2 θ + cos^2 θ - 2 = m^2    ...( v )

Now, Adding ( iv ) and ( v ) with the addition of 3 to it, and multiply by ( iii ),

= > ( iii )[ iv + v + 3 ]

= > ( sin^2θ.cos^2θ )( cosec^2θ + sin^2θ - 2 + sec^2θ + cos^2θ - 2 + 3 ) =  ( l^2 m^2 )( l^2 + m^2 + 3 )

= > ( sin^2θ.cos^2θ )( cosec^2θ + sec^2θ +sin^2θ + cos^2θ  - 4 + 3 ) = l^2 m^2( l^2 + m^2 + 3 )

= > ( sin^2θ.cos^2θ )( cosec^2θ + sec^2θ  + 1 - 4 + 3 ) = l^2m^2( l^2 + m^2 + 3 )

= > ( sin^2θ.cos^2θ )( cosec^2θ + sec^2θ ) = l^2m^2( l^2 + m^2 + 3 )

$\implies \sin^2\theta\times \cos^2\theta\times\bigg\{ \dfrac{1}{\sin^2\theta}+ \dfrac{1}{\cos^2\theta}\bigg\}\\\\\\\implies \sin^2\theta\times \cos^2\theta\times\bigg\{ \dfrac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\time\cos^2\theta}\bigg\}\\\\\\\implies\sin^2\theta\times\cos^2\theta \times \dfrac{1}{\sin^2\theta\times\cos^2\theta}\\\\\\\implies 1$

Hence, proved.