integral of (secx)(sinx)

Prove the following:[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
1. 3 answers
2. asked by Anonymous
3. 977 views
My previous question:Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) =
1. 2 answers
2. asked by Jon
3. 967 views
how do i simplify (secx - cosx) / sinx?i tried splitting the numerator up so that i had (secx / sinx) - (cosx / sinx) and then i
1. 1 answer
2. asked by v
3. 5,826 views
Simplify (1+tanx)^2The answer is (1-sinx)(1+sinx) Here's what I do: 1 + 2tanx + tan^2x When I simplify it becomes 1 +
1. 1 answer
2. asked by AlphaPrimes
3. 678 views
1/tanx-secx+ 1/tanx+secx=-2tanxso this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx)
1. 0 answers
2. asked by olivia
3. 1,373 views
Complete the following identity secx-1/secx=? I have four multiple choice options and can't seem to work my way to either one.
1. 3 answers
2. asked by Anne
3. 783 views
Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
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2. asked by Dave
3. 1,484 views
Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you!1. integral from 0 to pi/4 of
1. 0 answers
2. asked by Shelley
3. 806 views
That's the same as the integral of sin^2 x dx.Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx
1. 0 answers
2. asked by drwls
3. 873 views
I'm working this problem:∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx)
1. 1 answer
2. asked by Janice
3. 725 views