Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Verify the identity: (1-sinx)/cosx = cosx/(1+sinx)

See more answers at
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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

LS = (1-sinx)/cosx = 1/cosx - sinx/cosx RS = \(\large\frac{cosx}{1+sinx}\) = \(\large\frac{cosx(1-sinx)}{(1+sinx)(1-sinx)}\) = \( \large\frac{cosx}{1-sin^2x} - \frac{cosxsinx}{1-sin^2x}\) =.... Note: \[1 = cos^2x + sin^2x \]
why is it so tiny!
Because its a spoiler

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I'm very bad at basic math , so I'm finding all of this to be very foreign.
Actually, simply working from the left is ok :D \[RS = \frac{cosx}{(1+sinx)}\]\[=\frac{cos}{1+sinx}\times \frac{1-sinx}{1-sinx}\]By the identity a^2 - b^2 = (a+b)(a-b), we get\[=\frac{cos(1-sinx)}{1-sin^2x}\]Then, \(sin^2x+cos^2x =1\) to express \(1-sin^2x\) in terms of \(cos^2x\). Finally, cancel the common factor, and you'll get the left side!
so 1-sin^2 = cos^2x?
Oh, yes.
okay, thank you!
You're welcome~~

Not the answer you are looking for?

Search for more explanations.

Ask your own question