Tan^2x-sin^2x prove it can be written as tan^2xsin^2x?

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Answer 1 · 2017-11-13T15:37:52

See the proof below

We need

${tan}^{2} x + 1 = {sec}^{2} x$ $tan^2x+1=sec^2x$

$(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b)=a^2-b^2$

Therefore,

$L H S = {tan}^{2} x - {sin}^{2} x = (tan x + sin x) (tan x - sin x)$ $LHS=tan^2x-sin^2x=(tanx+sinx)(tanx-sinx)$

$= (\frac{sin x}{cos x} + sin x) (\frac{sin x}{cos x} - sin x)$ $=(sinx/cosx+sinx)(sinx/cosx-sinx)$

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

$= {sin}^{2} x (sec x + 1) (sec x - 1)$ $=sin^2x(secx+1)(secx-1)$

$= {sin}^{2} x ({sec}^{2} x - 1)$ $=sin^2x(sec^2x-1)$

$= {sin}^{2} x {tan}^{2} x$ $=sin^2xtan^2x$

$= R H S$ $=RHS$

$Q E D$ $QED$