If cot A + cosec A = 2 and A is an acute angle, then the value of $\frac{9 tan A+16 cosec A}{5sin A + 3 tan A}$ is :
Answer & explanation
Correct answer: option 3
cot A + cosec A = 2 ------(1)
we know, cosec²A - cot²A = 1
( cosecA - cotA ) ( cosecA + cotA ) = 1
( cosecA - cotA ) × 2 = 1
( cosecA - cotA ) = \(\frac{1}{2}\) ----(2)
On adding equation 1 and 2.
2 cosec A = 2 + \(\frac{1}{2}\)
cosec A = \(\frac{5}{4}\)
By using pythagoras theorem,
P² + B² = H²
4² + B² = 5²
B = 3
Now,
\(\frac{9tanA + 16 cosecA }{ 5sinA + 3 tanA }\)
= \(\frac{9×P/B + 16×H/P}{ 5×P/H + 3×P/B }\)
= \(\frac{9× 4/3 + 16×5/4}{ 5×4/5 + 3×4/3}\)
= \(\frac{12 + 20}{ 4 + 4}\)
= \(\frac{32}{8}\)
= 4