Answer :

We have, secθ = 5/4 = 1/cosθ

→ cosθ = (4k)/(5k) = AC/AB (For some value of k)

By Pythagoras theorem, (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}

∴AB^{2} = BC^{2} + AC^{2}

= (5k)^{2} = BC^{2} + (4k)^{2}

= 25k^{2} = BC^{2} + 16k^{2}

= BC^{2} = 9k^{2}

→ BC = 3k

∴ sinθ = BC/AB = (3k)/(5k) = 3/5

consider the LHS

LHS = =

=

=

=

=

= 12/7

= RHS

HENCE PROVED

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