Here are the solutions for your reference. In case of any queries do drop your questions so that it can be answered in time.

Ans 1 . *given 3x = cosecA, 3/x = cotA*

* (multiply and divide by 3 )*

* (identity 3)*

Ans 2 . *given 6x = secB , 6/x = tanB*

* ( multiply and divide by 36)*

* (substituting the values of 6x and 6/x)*

* ( identity 2 )*

* *

Ans 3 . Given

* *

* *

* *

substituting in eq(1).

* *

* *

* *

* *

* *

* *

* *

* *

* (identity 1)*

* *

Hence L.H.S = R.H.S

Ans 4.

* *

* *

* *

Ans 5.

Ans 6.

* ( identity 2 )*

Ans 7.

* √ *

Ans 8.

given tanA = p/q

LHS = (psinA – qcosA ) / (psinA + qcosA)

dividing both numerator and denominator by qcosA

hence proved.

Ans 9.

Given tan B = a/b

= (a sinB + bcosB) / (a sinB – bcosB)

dividing both numerator and denominator by bcosB

Ans 10.

given x = a Cos³A y = b Sin³A

( identity 1)