Prove the identity. tanx-cotx/tanx+cotx=sin^2x-cos^2x Have to show the statements and the rules?

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Prove the identity. tanx-cotx/tanx+cotx=sin^2x-cos^2x Have to show the statements and the rules?

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Answer 1 · 2018-04-07T11:51:34

See below.

Explanation:

Identities:

$tan x = \frac{sin x}{cos x}$ $color(red)bb(tanx=sinx/cosx)$

$cot x = \frac{cos x}{sin x}$ $color(red)bb(cotx=cosx/sinx)$

Substituting these in the LHS:

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

Adding the terms in numerator and denominator:

$\frac{\frac{{sin}^{2} x - {cos}^{2} x}{sin x cos x}}{\frac{{sin}^{2} x + {cos}^{2} x}{cos x sin x}}$ $((sin^2x-cos^2x)/(sinxcosx))/((sin^2x+cos^2x)/(cosxsinx)$

Multiplying by $(cos x sin x)$ $(cosxsinx)$

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

Identity:

${sin}^{2} x + {cos}^{2} x = 1$ $color(red)bb(sin^2x+cos^2x=1)$

Substituting this in the denominator:

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

${sin}^{2} x - {cos}^{2} x$ $sin^2x-cos^2x$

As required:

$L H S \equiv R H S$ $LHS-=RHS$

Answer 2 · 2018-04-07T15:10:43

$\frac{tan x - cot x}{tan x + cot x} = {sin}^{2} x - {cos}^{2} x$ $(tanx-cotx)/(tanx+cotx)=sin^2x-cos^2x$

$\Rightarrow \frac{tan x - \frac{1}{tan x}}{tan x + \frac{1}{tan x}}$ $=>(tanx-1/tanx)/(tanx+1/tanx)$

$\Rightarrow \frac{{tan}^{2} x - 1}{{tan}^{2} x + 1}$ $=>(tan^2x-1)/(tan^2x+1)$

$\Rightarrow \frac{{tan}^{2} x - 1}{{sec}^{2} x}$ $=>(tan^2x-1)/(sec^2x)$ $w w w w w w$ $color(white)(wwwwww$ $[as {tan}^{2} x - {sec}^{2} x = 1]$ $["as "tan^2x-sec^2x = 1]$

$\Rightarrow (\frac{{tan}^{2} x}{{sec}^{2} x} - \frac{1}{{sec}^{2} x})$ $=>(tan^2x/sec^2x-1/sec^2x)$

$\Rightarrow {sin}^{2} x - {cos}^{2} x$ $=>sin^2x-cos^2x$

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Prove the identity. tanx-cotx/tanx+cotx=sin^2x-cos^2x Have to show the statements and the rules?

2 Answers

Explanation:

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