Question #297e2

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Question #297e2

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Answer 1 · 2016-11-02T08:19:23

Both sides equals ${sin}^{2} t - {cos}^{2} t$ $sin^2t-cos^2t$

Explanation:

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

$\frac{tan t - cot t}{tan t + cot t} = \frac{\frac{sin t}{cos t} - \frac{cos t}{sin t}}{\frac{sin t}{cos t} + \frac{cos t}{sin t}} = \frac{\frac{{sin}^{2} - {cos}^{2} t}{sin t cos t}}{\frac{{sin}^{2} + {cos}^{2} t}{sin t cos t}} = {sin}^{2} t - {cos}^{2} t$ $(tant-cott)/(tant+cott)=(sint/cost-cost/sint)/(sint/cost+cost/sint)=((sin^2-cos^2t)/(sintcost))/((sin^2+cos^2t)/(sintcost))=sin^2t-cos^2t$

Answer 2 · 2016-11-06T01:22:39

$L H S = \frac{{tan}^{2} T - 1}{{sec}^{2} T}$ $LHS=(tan^2T-1)/sec^2T$

$= \frac{{tan}^{2} T - 1}{1 + {tan}^{2} T}$ $=(tan^2T-1)/(1+tan^2T)$

[Dividing numerator and denominator by $tan T$ $tanT$ ]

$= \frac{\frac{{tan}^{2} T}{tan T} - \frac{1}{tan T}}{\frac{1}{tan T} + \frac{{tan}^{2} T}{tan T}}$ $=(tan^2T/tanT-1/tanT)/(1/tanT+tan^2T/tanT)$

$= \frac{tan T - cot T}{cot T + tan T} = R H S$ $=(tanT-cotT)/(cotT+tanT)=RHS$

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Question #297e2

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