Question

inverse tigonometry question..........

Answer 1

You need help with the 15th question?

Answer 2

yes

Answer 3

Isolate x and y and then multiply them?

Answer 4

i solved them and also used the formula of sin(a + b) but i cannot simplify the expression which is given in the options

Answer 5

@phi

Answer 6

@Sachintha

Answer 7

this looks tricky. I don't see a fast way to the answer (hopefully there is one), but the answer is
\[ \sin^2 \theta + \cos^2 \beta \]

Answer 8

u are absolutely correct but i need explanation how u got that answer

Answer 9

"playing"... which is to say, I don't have a clear way to the solution
I expanded sin(A+B) to sinA cosB + cosA sin B
and sin(A-B) to sin A cos B - cosA sin B
multiplied to get sin^2 A cos^2 B - cos^2 A sin^2 B
so we have
1 + sin^2 A cos^2 B - cos^2 A sin^2 B
I decided to replace sin^2 B with (1- cos^2 B)
to get
1 + sin^2 A cos^2 B - cos^2 A (1- cos^2 B)

Answer 10

that mess becomes
1 + sin^2 A cos^2 B - cos^2 A + cos^2 A cos^2 B
or
1-cos^2 A + ( sin^2 A cos^2 B + cos^2 A cos^2 B)

sin^2 A + cos^2B ( sin^2 A+ cos^2 A)

sin^2 A + cos^2 B

Answer 11

but I keep thinking there must be a more "insightful" way to get the answer

Answer 12

thanx for the hint i got the answer..