Question #5a859

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Question #5a859

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Answer 1 · 2016-08-14T13:31:47

We will start from the left hand side and show that it equals the right hand side. To do so, we will use the following identities:

$tan(x) = sin(x)/cos(x)$
$sec(x) = 1/cos(x)$
$sec^2(x) - 1 = tan^2(x)$

Note that the third identity may be derived from $sin^2(x) + cos^2(x) = 1$ by dividing each side by $cos^2(x)$ and then subtracting $1$ from each side.

Proceeding:

$LHS = tan^2(u) - sin^2(u)$

$= (sin(u)/cos(u))^2-sin^2(u)$

$= sin^2(u)/cos^2(u)-sin^2(u)$

$=sin^2(u)(1/cos^2(u) - 1)$

$=sin^2(u)(sec^2(u)-1)$

$=sin^2(u)tan^2(u) = RHS$ ∎

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Question #5a859

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