anonymous
  • anonymous
Verify the identity.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ \cos7\theta + \cos5\theta }{ \sin7\theta - \sin5\theta } = \frac{ \csc \theta }{ \sec \theta } \]
Witty_Roger
  • Witty_Roger
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
anonymous
  • anonymous
Where did you get \[\sin7\theta - \cos5\] from?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Basically you wrote the equation like this, right? \[\frac{ \frac{ 2\cos(7\theta + 5\theta) }{ 2 } \times \frac{ \cos(7\theta - 5\theta) }{ 2 } }{ \ \frac{ 2\cos(7\theta + 5\theta) }{ 2 } \times \frac{ \sin(7\theta - 5\theta) }{ 2 }} = \frac{ 2\cos(6\theta)\cos \theta }{ 2\cos(6 \theta) \sin \theta }\]
Witty_Roger
  • Witty_Roger
yess
Witty_Roger
  • Witty_Roger
its a well established factorisation formula
Witty_Roger
  • Witty_Roger
sin A - sin B = 2 cos (A+B)/2 * sin(A-B)/2
anonymous
  • anonymous
Okay, but how did you get cscx/sinx if you originally has cosx/sinx, shouldn't it be secx/cscx?
Witty_Roger
  • Witty_Roger
sec x is inverse of cos x and cosec x is that of sin x
anonymous
  • anonymous
Yes, but didn't you have cosx/sinx, so shouldn't it be secx/cscx, not cscx/secx?
Witty_Roger
  • Witty_Roger
wait ill draw it n show u
Witty_Roger
  • Witty_Roger
|dw:1450251655023:dw|
Witty_Roger
  • Witty_Roger
satisfied??
anonymous
  • anonymous
Yeah, thanks.

Looking for something else?

Not the answer you are looking for? Search for more explanations.