Prove that $\frac{sin t - cos t}{sin t + cos t} = \frac{2 {sin}^{2} t - 1}{1 + 2 sin t cos t}$ $(sint-cost)/(sint+cost) = (2sin^2t-1)/(1+2sintcost)$ ?

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Prove that $\frac{sin t - cos t}{sin t + cos t} = \frac{2 {sin}^{2} t - 1}{1 + 2 sin t cos t}$ $(sint-cost)/(sint+cost) = (2sin^2t-1)/(1+2sintcost)$ ?

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Answer 1 · 2018-04-05T16:06:38

Well, 1 plays an important role in solving this problem.

Explanation:

Let us focus on the Left Hand Side (LHS)

$L H S = \frac{sin t - cos t}{sin t + cos t}$ $LHS=(sint -cost)/(sint+cost)$

Note that $1 = \frac{sin t + cos t}{sin t + cos t}$ $1 = (sint +cost)/(sint +cost)$

We will use this property to aid us because multiplying our LHS by 1 does not change its value or solutions.

That is,
$L H S = \frac{sin t - cos t}{sin t + cos t} \cdot \frac{sin t + cos t}{sin t + cos t}$ $LHS=(sint -cost)/(sint+cost)*(sint +cost)/(sint +cost$
Now we simplify, the numerator is a difference of two squares. Expand the numerator and rearranging terms, we have:

$L H S = \frac{{sin}^{2} t - {cos}^{2} t}{{(sin t + cos t)}^{2}}$ $LHS= (sin^2t-cos^2t)/(sint + cost)^2$

$L H S = \frac{2 {sin}^{2} t - 1}{{sin}^{2} t + {cos}^{2} t + 2 sin t cos t}$ $LHS= (2sin^2t - 1)/(sin^2t +cos^2t+2sintcost)$

We know that ${sin}^{2} t + {cos}^{2} t = 1$ $sin^2t+cos^2t=1$

So we have:

$L H S = \frac{2 {sin}^{2} t - 1}{1 + 2 sin t cos t}$ $LHS=(2sin^2t - 1)/(1+2sintcost)$

This is equal to the RHS. QED

Answer 2 · 2018-04-05T16:07:45

Please see below.

Explanation:

$\frac{sin t - cos t}{sin t + cos t}$ $(sint-cost)/(sint+cost)$

= $\frac{sin t - cos t}{sin t + cos t} \cdot \frac{sin t + cos t}{sin t + cos t}$ $(sint-cost)/(sint+cost)*(sint+cost)/(sint+cost)$

= $\frac{{sin}^{2} t - {cos}^{2} t}{{sin}^{2} t + {cos}^{2} t + 2 sin t cos t}$ $(sin^2t-cos^2t)/(sin^2t+cos^2t+2sintcost)$

= $\frac{{sin}^{2} t - (1 - {sin}^{2} t)}{1 + 2 sin t cos t}$ $(sin^2t-(1-sin^2t))/(1+2sintcost)$

= $\frac{{sin}^{2} t - 1 + {sin}^{2} t}{1 + 2 sin t cos t}$ $(sin^2t-1+sin^2t)/(1+2sintcost)$

= $\frac{2 {sin}^{2} t - 1}{1 + 2 sin t cos t}$ $(2sin^2t-1)/(1+2sintcost)$

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Prove that $\frac{sin t - cos t}{sin t + cos t} = \frac{2 {sin}^{2} t - 1}{1 + 2 sin t cos t}$ $(sint-cost)/(sint+cost) = (2sin^2t-1)/(1+2sintcost)$ ?

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Explanation:

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