Question

If cosec A/ sec B = p and cot A/ sec B = q, prove that:

Answer 1

\[(p^{2} - q^{2}) . \cot^{2}a= q ^{2}\]

Answer 2

This one seems pretty straight forward,
just plug everything in for the left side of the equation.

Answer 3

\[\large\rm \left(\left[\frac{\csc A}{\sec B}\right]^2-\left[\frac{\cot A}{\sec B}\right]^2\right)\cot^2A\]I replaced my p and q.
And then just simplify from there, ya?

Answer 4

I see

Answer 5

Using the identity, we get \[(\frac{ 1 }{ secB } )\cot ^{2}A\]

Answer 6

Yea?

Answer 7

Hmm the denominators should be squared as well, ya?\[\large\rm \left(\frac{\csc^2A}{\sec^2B}-\frac{\cot^2A}{\sec^2B}\right)\cot^2A\]

Answer 8

\[\large\rm \left(\frac{\csc^2A-\cot^2A}{\sec^2B}\right)\cot^2A\]But yes, your Pythagorean Identity cleans up the numerator,\[\large\rm \left(\frac{1}{\sec^2B}\right)\cot^2A\]

Answer 9

Dang, didn't notice that ;-;

Answer 10

Haha I love how you called it "the" identity.
We're in trig XD the land of identities.
But I knew which one you were talking about :D lol

Answer 11

XD I suck at trig btw ;-;

Answer 12

So, it's always 'the' identity :P

Answer 13

So now we get \[(\frac{ \cot ^{2}A }{ \sec ^{2} B }\]

Answer 14

Forgot to close the bracket -.-

Answer 15

k looks good.

Answer 16

Yea?