Question

prove that,sec 4 theetha-1 / sec 2 theetha-1= tan 4 theetha/tan theetha

Answer 1

Prove that :-

$\rm \: \: \: \: \: \: \: \: \: \: \dfrac{sec4\theta - 1}{sec2\theta - 1} = \dfrac{tan4\theta }{tan\theta }$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: \dfrac{sec4\theta - 1}{sec2\theta - 1}$

can be rewritten as

$\rm \: =  \: \dfrac{\dfrac{1}{cos4\theta } - 1}{\dfrac{1}{cos2\theta } - 1}$

$\rm \: =  \: \dfrac{\dfrac{1 - cos4\theta }{cos4\theta }}{\dfrac{1 - cos2\theta }{cos2\theta }}$

can be rewritten as

$\rm \: =  \: \dfrac{(1 - cos4\theta )cos2\theta }{(1 - cos2\theta) cos4\theta }$

We know,

$\boxed{\sf{  \:1 - cos2x = {2sin}^{2}x \: }}$

So, using this result, we get

$\rm \: =  \: \dfrac{2 {sin}^{2} 2\theta \times cos2\theta }{2 {sin}^{2} \theta \times cos4\theta }$

$\rm \: =  \: \dfrac{2 {sin}2\theta \times sin2\theta \times cos2\theta }{2 {sin}^{2} \theta \times cos4\theta }$

can be further rearrange as

$\rm \: =  \: \dfrac{{sin}2\theta \times(2 sin2\theta cos2\theta )}{2 {sin}^{2} \theta \times cos4\theta }$

We know,

$\boxed{\rm{  \:2 \: sinx \: cosx \: = \: sin2x \: }}$

So, using this result, we get

$\rm \: =  \: \dfrac{(2sin\theta cos\theta ) \times sin4\theta }{2 {sin}^{2} \theta \times cos4\theta }$

$\rm \: =  \: \dfrac{ cos\theta\times sin4\theta }{sin\theta \times cos4\theta }$

$\rm \: =  \: \dfrac{tan4\theta }{tan\theta }$

Hence,

$\rm\implies \:\rm \: \: \: \boxed{\rm{  \: \: \: \: \: \: \: \: \dfrac{sec4\theta - 1}{sec2\theta - 1} = \dfrac{tan4\theta }{tan\theta } \: \: \: \: }}$

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Additional Information :-

$\boxed{\sf{  \:cos2x = {2cos}^{2}x - 1 \: }}$

$\boxed{\sf{  \:cos2x = 1 - {2sin}^{2}x \: }}$

$\boxed{\sf{  \:cos2x = {cos}^{2}x - {sin}^{2}x \: }}$

$\boxed{\sf{  \:cos2x = \frac{1 - {tan}^{2}x }{1 + {tan}^{2}x } \: }}$

$\boxed{\sf{  \:sin2x = \frac{2tanx }{1 + {tan}^{2}x } \: }}$

$\boxed{\sf{  \:tan2x = \frac{2tanx }{1 - {tan}^{2}x } \: }}$

$\boxed{\sf{  \:sin3x = 3sinx - {4sin}^{3}x \: }}$

$\boxed{\sf{  \:cos3x = {4cos}^{3}x - 3cosx \: }}$

$\boxed{\sf{  \:tan3x = \frac{3tanx - {tan}^{3} x}{1 - {3tan}^{2} x} \: }}$