Question

Question no I've prove the identity

Answer 1

Answer :

Let us know about a few trigonometric identity formulae which should be well-known.

sin²θ + cos²θ = 1 ...(i)

sec²θ - tan²θ = 1 ...(ii)

cosec²θ - cot²θ = 1 ...(iii)

Algebraic identity -

a² - b² = (a + b) (a - b) ...(iv)

Now, L.H.S.

= cot²θ/(cosecθ + 1)

= (cosec²θ - 1)/(cosecθ + 1), using (iii)

= [(cosecθ)² - 1²]/(cosecθ + 1)

= [(cosecθ + 1) (cosecθ - 1)]/(cosecθ + 1), using (iv)

= cosecθ - 1

= R.H.S. [Proved]

Hope it helps!

Answer 2

Here is your solution :

( iv )

R.H.S = cosec ∅ - 1

L.H.S = cot²∅ / ( cosec∅ + 1 )
_________________________

We know that,

=> cot∅ = cos∅ / sin∅

Squaring both sides,

=> cot²∅ = cos²∅ / sin²∅

And,

=> cosec∅ = ( 1 / sin∅ )
_________________________

= ( cos²∅ / sin²∅ ) / [ ( 1 / sin∅ ) + 1]

= ( cos²∅ / sin²∅ ) / [ ( 1 + sin∅ ) / sin∅ ]

= ( cos²∅ / sin²∅ ) × [ sin∅ / ( 1 + sin∅ ) ]

= cos²∅ / [ sin∅ ( 1 + sin∅ ) ]

= cos²∅ / [ sin∅ ( 1 + sin∅ ) ]

We know that,

=> sin²∅ + cos²∅ = 1

•°• cos²∅ = 1 - sin²∅

= ( 1 - sin²∅ ) / [ sin∅ ( 1 + sin∅) ]

= [ 1² - ( sin∅ )²] / [ sin∅ ( 1 + sin∅ ) ]

Using identity,

=> ( a² - b² ) = ( a + b ) ( a - b )

= [ ( 1 + sin∅ ) ( 1 - sin∅ ) ] / [ sin∅( 1 + sin∅ ) ]

= ( 1 - sin∅ ) / sin∅

= ( 1 / sin∅ ) - ( sin∅ / sin∅ )

We know that,

=> sin∅ = 1 / cosec∅

•°• cosec∅ = 1 / sin∅

= cosec∅ - 1 ( R.H.S )

Proved

Hope it helps !!