The correct option is B tan 8Atan 2A Given, 8A 1 4A 1 = (1/cos8A) 1) (1/cos4A) 1 =(1 cos8A)/cos8A) (1 cos4A) / cos4A) = cos4 ...

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Trigonometric Ratios

sec 8A-1sec 4...

Question

Evaluate: $\frac{sec 8 A - 1}{sec 4 A - 1} =$

0

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\frac{t a n 8 A}{t a n 2 A}

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\frac{c o s 8 A}{c o s 2 A}

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\frac{s i n 8 A}{s i n 2 A}

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Solution

The correct option is B $\frac{tan 8 A}{tan 2 A}$
Given,
$\frac{sec 8 A - 1}{sec 4 A - 1}$
$= \frac{\frac{1}{cos 8 A} - 1}{\frac{1}{cos 4 A} - 1}$
$= \frac{\frac{1 - cos 8 A}{cos 8 A}}{\frac{1 - cos 4 A}{cos 4 A}}$
$= \frac{cos 4 A (1 - cos 8 a)}{cos 8 A (1 - cos 4 A)}$
$= \frac{cos 4 A (1 - (1 - 2 {sin}^{2} 4 A))}{cos 8 A (1 - (1 - 2 {sin}^{2} 2 A))}$ [ $∵ cos 2 θ = 1 - 2 {sin}^{2} θ$ ]
$= \frac{cos 4 A (2 {sin}^{2} 4 A)}{cos 8 A (2 {sin}^{2} 2 A)}$
$= \frac{(2 cos 4 A sin 4 A) sin 4 A}{(2 cos 8 A {sin}^{2} 2 A)}$ [ $∵ sin 2 θ = 2 sin θ cos θ$ ]
$= \frac{sin 8 A sin 4 A}{(2 cos 8 A {sin}^{2} 2 A)}$
$= \frac{tan 8 A sin 4 A}{2 {sin}^{2} 2 A}$
$= \frac{tan 8 A (2 sin 2 A cos 2 A)}{2 {sin}^{2} 2 A}$
$= \frac{tan 8 A cos 2 A}{sin 2 A}$
$= \frac{tan 8 A}{tan 2 A}$

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Similar questions

A = 30^{0}

cos 2 A = {cos}^{2} A - {sin}^{2} A = \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A}

{sin}^{8} A - {cos}^{8} A = ({sin}^{2} A - {cos}^{2} A) (1 - 2 {sin}^{2} A {cos}^{2} A)

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $s i n^{2} A - c o s^{2} A$

3.tan A = $\frac{2 t a n \frac{A}{2}}{1 - t a n^{2} \frac{A}{2}}$

4. $s i n^{2} \frac{A}{2}$ = 1 - cosA

5. $t a n^{2} A$ = $\frac{1 - c o s 2 A}{1 + c o s 2 A}$

6. $s i n^{2} A$ = $\frac{2 t a n A}{1 - t a n^{2} A}$

\frac{{cos}^{2} A + {tan}^{2} A - 1}{{sin}^{2} A} = {tan}^{2} A .

30^{\circ}

cos 2 A = \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A} = {cos}^{2} A - {sin}^{2} A

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