Question

plzzz solve questions no.(c).

Answer 1

given
cosA = 12/5 sinA ______(1
we have to prove
Tan^2- sin^2A = sin^4A.sec^2A ____(2
From eqn (1
cosA/sinA = 12/5 = tanA
and SinA = 5/12(cosA)
put these values in eqn (2
L.H.S =(12/5)^2 - (5/12cosA)^2
= 144/25 - 25/144(cos^2A)
= 144/25(1-cos^2A) = 144/25(sin^2A)
R.H.S = (5/12cosA)^4 × sec^2A
= 25/144.cos^4A × sec^2A
= 25/144 cos^4A × 1/cos^2A
= 25/144.cos^2A
hence L.H.S = R.H.S
Proved

Answer 2

Given 5 cos A = 12 sin A

           5 cos A = 12 sin A   ------------ (1)

tan A = sin A/cos A

          = 5 cos A/12/cos A

          = 5/12.


We know that sec^2 A = 1 + tan^2 A

                                       = 1 + (5/12)^2

                                       = 1 + 25/144

                                        = 169/144

                               sec A = 13/12.


We know that cos A = 1/sec A

                                   = 1/13/12

                                   = 12/13.


We know that sin^2 A = 1 - cos^2 A

                                      = 1 - (12/13)^2

                                      = 1 - 144/169

                                      = 25/169

                             sin A = 5/13.



LHS = tan^2 A - sin^2 A

        = (5/12)^2 - (5/13)^2

        = 5^2/12^2 - 5^2/169

        = 5^2 * 169 - 5^2 * 144/12^2 * 13^2

        = 25 * 169 - 25 * 144/24336

        = 4225 - 3600/24336

        = 625/24336.



RHS = sin^4 A * sec^2 A

        = (5/13)^4 * (13/12)^2

        = 5^4/13^4 * 13^2/12^2

        = 5^4/13^2 * 12^2

        = 625/169 * 144

        = 625/24336.



LHS = RHS.



Hope this helps!