Question

If sinA=-5/13 and A is in the third quadrant, then prove that 5cot^2A+12tanA+13cosecA=0

Answer 1

Answer:

....

Step-by-step explanation:

By Pythagoras' theorem, you can find value of tan(A)=5\12 and it will be positive as A is in 3rd quadrant.

Also, $cot^2A = cosec^2A -1$

So the LHS is:

$=5(cosec^2A -1) + 12tanA +13cosecA\\=5(\frac{169}{25} -1)+12(\frac{5}{12}) + 13(\frac{-13}{5})\\=5(\frac{144}{25}) + 5 - \frac{169}{5}\\=\frac{144}{5} - \frac{144}{5}\\=0$

=RHS