Question

Verify the identity.

Answer 1

\[\frac{ \cos7\theta + \cos5\theta }{ \sin7\theta - \sin5\theta } = \frac{ \csc \theta }{ \sec \theta } \]

Answer 2

cos 7@+cos5@ =2 cos[(7@+5@)/2] * cos[(7@-5@)/2] = 2 cos6@ cos@
sin 7@ - cos 5@ = 2 cos[(7@+5@)/2] * sin[(7@-5@)/2] = 2 cos6@ sin@
so after substituting u get
LHS = cos@/sin@ =cosec@/sec@=RHS
Hence proved

Answer 3

Where did you get \[\sin7\theta - \cos5\] from?

Answer 4

Basically you wrote the equation like this, right?

\[\frac{ \frac{ 2\cos(7\theta + 5\theta) }{ 2 } \times \frac{ \cos(7\theta - 5\theta) }{ 2 } }{ \ \frac{ 2\cos(7\theta + 5\theta) }{ 2 } \times \frac{ \sin(7\theta - 5\theta) }{ 2 }} = \frac{ 2\cos(6\theta)\cos \theta }{ 2\cos(6 \theta) \sin \theta }\]

Answer 5

yess

Answer 6

its a well established factorisation formula

Answer 7

sin A - sin B = 2 cos (A+B)/2 * sin(A-B)/2

Answer 8

Okay, but how did you get cscx/sinx if you originally has cosx/sinx, shouldn't it be secx/cscx?

Answer 9

sec x is inverse of cos x and
cosec x is that of sin x

Answer 10

Yes, but didn't you have cosx/sinx, so shouldn't it be secx/cscx, not cscx/secx?

Answer 11

wait ill draw it n show u

Answer 12

satisfied??

Answer 13

Yeah, thanks.