Question

please answer this with proper steps.

Answer 1

Answer:

Step-by-step explanation:

(tanФ+sinФ)²-(tanФ-sinФ)²

=4×tan²Ф×sin²Ф

=4√(tanФ+sinФ)(tanФ-sinФ)

=4√mn

Answer 2

TRIGONOMETRY :-

IT is given that :

tanθ+sinθ=m

tanθ-sinθ=n

so from both eqn ,

(m+n) = tanθ+sinθ+tanθ-sinθ

          =2tanθ (+sinθ - sinθ gets eliminated )-----eqn1

(m-n) = tanθ+sinθ-(tanθ-sinθ)

(m-n) =  (tanθ+sinθ-tanθ+sinθ)

         =2sinθ( +tan-tanθ gets cancelled ) -------eqn2

NOW find mn

mn = (tanθ+sinθ)(tanθ-sinθ)

    =tan²θ-sin²θ (they are in the from of (a+b)(a-b)-------eqn3

Now it is given to prove  = m²-n²= 4√mn

For m²-n² we have

=(m+n)(m-n)

=2tanθ.2sinθ (From eqn 1 qnad 2 )

= 4sinθtanθ (LHS)

NOW for  = 4√mn,we have

=4√(tan²θ-sin²θ)   [from eqn 3]

simplify the term

=4√(sin²θ/cos²θ-sin²θ)

=4√sin²θ(1/cos²θ-1)

=4sinθ√(1-cos²θ)/cos²θ

=4sinθ/cosθ√sin²θ [∵, sin²θ+cos²θ=1]

=4sinθtanθ (RHS)

∴ From above the LHS = RHS and hence it is proved

HOPE this helped you +_+