Prove it: ${tan}^{5} x = \frac{(\frac{1}{{(1 - sin x)}^{2}}) - (\frac{1}{{(1 + sin x)}^{2}})}{(\frac{1}{{(1 - cos x)}^{2}}) - (\frac{1}{{(1 + cos x)}^{2}})}$ $tan^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)$ ?

Question

Prove it: ${tan}^{5} x = \frac{(\frac{1}{{(1 - sin x)}^{2}}) - (\frac{1}{{(1 + sin x)}^{2}})}{(\frac{1}{{(1 - cos x)}^{2}}) - (\frac{1}{{(1 + cos x)}^{2}})}$ $tan^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)$ ?

2 Answers

Impact of this question

1696 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-16T20:04:12

To prove

$t g^{5} x = \frac{(\frac{1}{{(1 - sin x)}^{2}}) - (\frac{1}{{(1 + sin x)}^{2}})}{(\frac{1}{{(1 - cos x)}^{2}}) - (\frac{1}{{(1 + cos x)}^{2}})}$ $tg^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)$

RHS

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

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

$= \frac{\frac{4 sin x}{{cos}^{4} x}}{\frac{4 cos x}{{sin}^{4} x}}$ $=((4sinx)/cos^4x)/((4cosx)/(sin^4x))$

$= \frac{{sin}^{5} x}{{cos}^{5} x} = {tan}^{5} x = L H S$ $=sin^5x/cos^5x=tan^5x=LHS$

Proved

Answer 2 · 2017-03-16T20:28:59

This is one of those proofs that is easier to work from right to left. Start with:

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

Multiply numerator and denominator of the embedded fractions by the "conjugates" (e.g. $1 \pm sin x$ $1pmsinx$ on $1 \mp sin x$ $1 ∓ sinx$ ). You get that, e.g., $(1 + sin x) (1 - sin x) = 1 - {sin}^{2} x$ $(1+sinx)(1-sinx) = 1-sin^2x$ .

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

Repeat the previous step to simplify the denominator in the embedded fractions further:

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

Use the identities $1 - {sin}^{2} x = {cos}^{2} x$ $1-sin^2x = cos^2x$ and $1 - {cos}^{2} x = {sin}^{2} x$ $1-cos^2x = sin^2x$ to get:

$= \frac{(\frac{{(1 + sin x)}^{2}}{{cos}^{4} x}) - (\frac{{(1 - sin x)}^{2}}{{cos}^{4} x})}{(\frac{{(1 + cos x)}^{2}}{{sin}^{4} x}) - (\frac{{(1 - cos x)}^{2}}{{sin}^{4} x})}$ $= (((1+sinx)^2/(cos^4x))-((1-sinx)^2/(cos^4x)))/(((1+cosx)^2/(sin^4x))-((1-cosx)^2/(sin^4x))$

Combine fractions and flip to multiply the reciprocals:

$= \frac{\frac{{(1 + sin x)}^{2} - {(1 - sin x)}^{2}}{{cos}^{4} x}}{\frac{{(1 + cos x)}^{2} - {(1 - cos x)}^{2}}{{sin}^{4} x}}$ $= (((1+sinx)^2-(1-sinx)^2)/(cos^4x))/(((1+cosx)^2-(1-cosx)^2)/(sin^4x))$

$= \frac{{(1 + sin x)}^{2} - {(1 - sin x)}^{2}}{{cos}^{4} x} \cdot \frac{{sin}^{4} x}{{(1 + cos x)}^{2} - {(1 - cos x)}^{2}}$ $= ((1+sinx)^2-(1-sinx)^2)/(cos^4x)*(sin^4x)/((1+cosx)^2-(1-cosx)^2)$

Expand the squared terms:

$= (cancel(1)+2sinx+cancel(sin^2x)-(cancel(1)-2sinx+cancel(sin^2x)))/(cos^4x)*(sin^4x)/(cancel(1)+2cosx+cancel(cos^2x)-(cancel(1)-2cosx+cancel(cos^2x)))$

$= (cancel(4)sinx)/(cos^4x)*(sin^4x)/(cancel(4)cosx)$

$= color(blue)(tan^5x)$

Science

Math

Humanities

... and beyond

Prove it: ${tan}^{5} x = \frac{(\frac{1}{{(1 - sin x)}^{2}}) - (\frac{1}{{(1 + sin x)}^{2}})}{(\frac{1}{{(1 - cos x)}^{2}}) - (\frac{1}{{(1 + cos x)}^{2}})}$ $tan^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)$ ?

2 Answers

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Prove it: tan^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)tan5x=(1(1−sinx)2)−(1(1+sinx)2)(1(1−cosx)2)−(1(1+cosx)2)tan^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)?

2 Answers

Related questions

Impact of this question

Prove it: ${tan}^{5} x = \frac{(\frac{1}{{(1 - sin x)}^{2}}) - (\frac{1}{{(1 + sin x)}^{2}})}{(\frac{1}{{(1 - cos x)}^{2}}) - (\frac{1}{{(1 + cos x)}^{2}})}$ $tan^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)$ ?