how do i prove this????? pls help - tan2x + sec2x = 1

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

how do i prove this????? pls help - tan2x + sec2x = 1

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

-tan2x+sec2x \[\frac{ -\sin2x }{ \cos2x }+\frac{ 1 }{ \cos2x }\]\[\frac{ -\sin2x+1 }{ \cos2x }\]\[1-\sin2x=\cos2x\]so\[\frac{ \cos2x }{ \cos2x }=1\]..proven...
thank you sooo much @LynFran
ur welcome..please medal..thanks

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@DecentNabeel its the Pythagorean identity\[cosx+sinx=1\]so \[\cos2x+\sin2x=1 \] so \[\cos2x=1-\sin2x\]
@freckles please tell @DecentNabeel im right
I'm gonna have to agree with @DecentNabeel here but I do think he meant to write \[-\tan^2(x)+\sec^2(x)=1 \text{ instead of } -\tan(2x)+\sec(2x)=1\]
that last I was thinking about the OP when I said he
\[-\tan ^2x+\sec ^2x=1\]
Is that^^^what you are trying to prove?
\[\sin(x)+\cos(x) \neq 1 \text{ for all } x \\ \text{ the pythagorean identity is } \\ \sin^2(x)+\cos^2(x)=1\] |dw:1435970814153:dw| we see here the following: \[\cos(\theta)=\frac{x}{r} \implies r \cos(\theta)= x \\ \sin(\theta)=\frac{y}{r} \implies r \sin(\theta)= y \\ \text{ now by Pythagorean theorem we have } \\ x^2+y^2=r^2 \\(r \cos(\theta))^2+(r \sin(\theta))^2=r^2 \\ r^2 \cos^2(\theta)+r^2 \sin^2(\theta)=r^2 \\ \text{ divide both sides by } r^2 \\ \cos^2(\theta)+\sin^2(\theta)=1 \]
I see that the asker has disappeared.

Not the answer you are looking for?

Search for more explanations.

Ask your own question