Question

Prove the following trigonometry question

$\frac{sec8x - 1}{sec4x - 1} = \frac{tan8x}{tan2x}$
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Answer 1

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

Consider LHS

$\rm \: \dfrac{sec8x - 1}{sec4x - 1}$

can be rewritten as

$\rm \: = \dfrac{ \dfrac{1}{cos8x} - 1}{ \dfrac{1}{cos4x} - 1}$

$\rm \: = \dfrac{ \dfrac{1 - cos8x}{cos8x}}{ \dfrac{1 - cos4x}{cos4x}}$

$\rm \: = \dfrac{1 - cos8x}{1 - cos4x} \times \dfrac{cos4x}{cos8x}$

We know,

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

So, using this result, we get

$\rm \: = \dfrac{ {2sin}^{2} 4x}{ {2sin}^{2} 2x} \times \dfrac{cos4x}{cos8x}$

$\rm \: = \: \dfrac{2sin4x \times sin4x \times cos4x}{2sin2x \times sin2x \times cos8x}$

can be re-arranged as

$\rm \: = \: \dfrac{2sin4xcos4x \times sin4x}{2sin2x \times sin2x \times cos8x}$

We know,

$\boxed{ \rm{ \:2sinxcosx = sin2x \: \: }}$

So, using this result, we get

$\rm \: = \dfrac{sin8x \times (2sin2xcos2x)}{2sin2x \times sin2x \times cos8x}$

$\rm \: = \dfrac{sin8x \times cos2x}{sin2x \times cos8x}$

$\rm \: = \dfrac{sin8x}{cos8x} \times \dfrac{cos2x}{sin2x}$

$\rm \: = \: tan8x \times cot2x$

$\rm \: = \dfrac{tan8x}{tan2x}$

Hence,

$\rm\implies \:\rm \: \boxed{ \rm{ \:\dfrac{sec8x - 1}{sec4x - 1} = \frac{tan8x}{tan2x} \: \: }}$

$\rule{190pt}{2pt}$

$\begin{gathered} { \boxed{ \begin{array}{c} \underline{\underline{ \color{orange} \text{Additional \: lnformation}}} \\& \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered}$

Answer 2

Answer:

Step-by-step explanation:

Consider LHS ,

As we know sec θ = 1/cos θ

So,

As we know 1 – cos 2x = 2 sin2x

So,

As we know sin 2x = 2 sin x cos x

So,

= tan 8x cot 2x

= RHS

Hence proved