Question

sinA=$\frac{12}{13}$    0degree < A < 90degree   find the value of 
$\frac{sin²-cos²}{2 sin.cos}. \frac{1}{tan²}$

Answer 1

first we draw a right triangle with any one angle as A...then we substitute the values of sinA=opposite side to angle A/hypotenuse=12/13
We then give the values to the hypotenuse as 13k (for some common factor k for 13 and 12) and the side opposite to angle A as 12k,
then we apply Pythagoras theorem to the triangle to find the value of the 3rd side in the triangle
we get that the value of the side adjacent to angle A is 5k.
Now we find the values of cosA and tanA
we get cosA=5/13 and tan A = 12/5
now we substitute these values in the equation to find the answer......
we get:
(144/169-25/169÷2×12/13×5/13)×5/12
=119/120×5/12= 119/288 = 0.413