Asked by Kevin
Tan theta over 1 + sec theta + 1 +sec theta over Tan theta= 2 csc theta
Answers
Answered by
Kevin
Verify the identity:
Answered by
Reiny
totally confusing the way you typed.
Lucky that I know the relation is supposed to say:
tanθ/(1+secθ) + (1+secθ)/tanθ = 2cscθ
LS = (tan^2 θ + (1+secθ)^2)/(tanθ(1+secθ)) , using a common denominator
= (tan^2 θ + 1 + 2secθ + sec^2 θ)((tanθ(1+secθ))
= (sec^2 θ - 1 + 1 + 2secθ + sec^2 θ)/(tanθ(1+secθ))
= 2secθ(secθ + 1)/(tanθ(1+secθ))
= 2secθ/tanθ , secθ ≠ -1
= 2(1/cosθ)(cosθ/sinθ)
= 2/sinθ
= 2cscθ
= RS
Lucky that I know the relation is supposed to say:
tanθ/(1+secθ) + (1+secθ)/tanθ = 2cscθ
LS = (tan^2 θ + (1+secθ)^2)/(tanθ(1+secθ)) , using a common denominator
= (tan^2 θ + 1 + 2secθ + sec^2 θ)((tanθ(1+secθ))
= (sec^2 θ - 1 + 1 + 2secθ + sec^2 θ)/(tanθ(1+secθ))
= 2secθ(secθ + 1)/(tanθ(1+secθ))
= 2secθ/tanθ , secθ ≠ -1
= 2(1/cosθ)(cosθ/sinθ)
= 2/sinθ
= 2cscθ
= RS
Answered by
oobleck
extra credit: During Reiny's proof, he came to the step where he noted
= 2secθ/tanθ , secθ ≠ -1
Are there any other values you need to watch out for ?
= 2secθ/tanθ , secθ ≠ -1
Are there any other values you need to watch out for ?
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