Prove that tan^2 x+1=sec^2x?

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Answer 1 · 2017-07-12T16:18:14

See the proof below

We need

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

$sec x = \frac{1}{cos x}$ $secx=1/cosx$

Therefore,

$L H S = {tan}^{2} x + 1$ $LHS=tan^2x+1$

$= \frac{{sin}^{2} x}{{cos}^{2} x} + 1$ $=sin^2x/cos^2x+1$

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

$= \frac{1}{{cos}^{2} x}$ $=1/cos^2x$

$= {sec}^{2} x$ $=sec^2x$

$= R H S$ $=RHS$

$Q E D$ $QED$