The value of $\frac{3tan^260°+sec^230°-sin^245°}{(cos15°+sin75°)(sec15°+cosec75°)}$ is :
Answer & explanation
Correct answer: option 4
$\frac{3tan^260°+sec^230°-sin^245°}{(cos15°+sin75°)(sec15°+cosec75°)}$
{ using , sinA = cos ( 90° - A ) }
= $\frac{3 ×(√3)²+(2/√3)² - (1/√2)²}{(cos15°+cos15°)(sec15°+sec15°)}$
= $\frac{9+4/3 - 1/2}{(2cos15°)(2sec15°)}$
= $\frac{(54+8 - 3 )}{6}$
= $\frac{(59 )}{24}$