Question

solve it fast class 10 trigo.
Try to solve without actually finding the values

Answer 1

Here is your solution :

Given,

=> cot∅ = ( 1 / √3 )

=> R.H.S = ( 3/5 )


=> L.H.S = ( 1 - cos²∅ ) / ( 2 - sin²∅ ) -- ( 1 )

We know that,

=> sin∅ / cos∅ = tan∅

Using identity ,

=> tan∅ = ( 1/cot∅ )

=> sin∅ / cos∅ = ( 1 / cot∅ )

=> cot∅ = cos∅ / sin∅

=> cos∅ = cot∅ × sin∅

Using identity ,

=> sin∅ = ( 1/cosec∅ )


=> cos∅ = cot∅ ( 1/cosec∅ )

=> cos∅ = cot∅ / cosec∅

Using identity,

=> cosec²∅ - cot²∅ = 1

=> cosec²∅ = ( 1 + cot²∅ )

=> cosec∅ = √( 1 + cot²∅ )


=> cos∅ = cot∅ / √( 1 + cot² ∅ ) ------ ( 2 )

Again,


We know that,

=> cosec²∅ - cot²∅ = 1

=> cosec²∅ = ( 1 + cot²∅ )

Using identity,

=> cosec∅ = ( 1/sin∅ )

=> ( 1/sin∅ )² = ( 1 + cot²∅ )

=> 1 / sin²∅ = ( 1 + cot²∅ )

=> sin²∅ = 1 / ( 1 + cot²∅ )

=> sin∅ = √[ 1 / ( 1 + cot²∅ ) ]

•°• sin∅ = 1 / √( 1 + cot²∅ ). ----------- ( 3 )

Substitute the value of ( 2 ) and ( 3 ) in ( 1 ),

= ( 1 - cos²∅ ) / ( 2 - sin²∅ )

= [ 1 - ( cos∅ )² ] / [ 2 - ( sin∅ )² ]

= [ 1 - { cot∅/√( 1 + cot²∅ ) }² ] / [ 2 - { 1/√( 1 + cot²∅ )}² ]

= [ 1 - { cot²∅ / ( 1 + cot²∅ ) } ] / [ 2 - { 1 / ( 1 + cot²∅ ]

Numerator :

= [ 1 - { cot²∅ / ( 1 + cot²∅ } ]

= ( 1 + cot²∅ - cot²∅ ) / ( 1 + cot²∅ )

= 1 / ( 1 + cot²∅ )

Denominator :

= [ 2 - { 1 / ( 1 + cot²∅ ) } ]


= [ 2 + 2 cot²∅ - 1 ] / ( 1 + cot²∅ )

= ( 1 + 2 cot²∅ ) / ( 1 + cot²∅ )

Fraction = Numerator / Denominator

= [ 1 / ( 1 + cot²∅ ) ] / [ ( 1 + 2 cot²∅ ) / ( 1 + cot²∅ ) ]


= [ 1 / ( 1 + cot²∅ ) ] × [ ( 1 + cot²∅ ) / ( 1 + 2 cot²∅ ) ]


= 1 / ( 1 + 2 cot²∅ )

Substitute the value of cot∅,

= 1 / [ 1 + 2 ( 1 / √3 )² ]

= 1 / [ 1 + 2 ( 1 / 3 ) ]

= 1 / [ ( 3 + 2 ) / 3 ]

= 1 / [ 5 / 3 ]

=> ( 3 / 5 ) = R.H.S