Question

Please answer the question............

Answer 1

Note : Theta is written as A.

Answer:

Required numeric value of cosecA + cotA is √2 + 1.

Step-by-step explanation:

Given,

cosecA - cotA = √2 - 1

From the properties of trigonometry :

• cosec^2 B - cot^2 B = 1 , where B is an angle.

Thus,

= > cosec^2 A - cot^2 A = 1

= > ( cosecA + cotA )( cosecA - cotA ) = 1 { using a^2 - b^2 = ( a + b ) ( a - b) }

= > ( cosecA + cotA )( √2 - 1 ) = 1 { given, cosecA - cotA = √2 - 1 }

= > cosecA + cotA = 1 / ( √2 - 1 )

On right hand side, divide and multiply by √2 + 1 :

= > cosecA + cotA = 1 / ( √2 - 1 ) x ( √2 + 1 ) / ( √2 + 1 )

= > cosecA + cotA = ( √2 + 1 ) / ( √2 - 1 )( √2 + 1 )

= > cosecA + cotA = ( √2 + 1 ) / { ( √2 )^2 - 1^2 } { using ( a + b )( a - b ) = a^2 - b^2 }

= > cosecA + cotA = ( √2 + 1 ) / ( 2 - 1 ) = ( √2 + 1 ) / 1

= > cosecA + cotA = √2 + 1

Hence the required numeric value of cosecA + cotA is √2 + 1.

Answer 2

Solution :-

Given

cosec θ - cot θ = √2 - 1

We know that

cosec² θ - cot² θ = 1

It can be written as

⇒ (cosec θ + cot θ)(cosec θ - cot θ) = 1

Since a² - b² = (a + b)(a - b)

By substituting the value of cosec θ - cot θ in the above equation

⇒ (cosec θ + cot θ)(√2 - 1) = 1

⇒ cosec θ + cot θ = 1/(√2 - 1)

By rationalising the denominator of RHS

⇒ cosec θ + cot θ = 1/(√2 - 1) * (√2 + 1)/(√2 + 1)

⇒ cosec θ + cot θ = (√2 + 1)/(2 - 1)

⇒ cosec θ + cot θ = (√2 + 1)/1

⇒ cosec θ + cot θ = √2 + 1

Therefore the value of cosec θ + cot θ is √2 + 1