8 A -1- 4 A -1 is equal toA- tan 8 A -tan 2 AB- tan 2-tan 8 AD- none of these

The correct option is A tan 8A/tan 2A We have sec 8A 1/sec 4A 1=1/cos 8A 1/1/cos 4A 1 =cos 4A/cos 8A×1 cos 8A/1 cos 4A = cos 4A/cos 8A×2 sin^2 4A/2 sin^2 2A ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

8 A -1/ 4 A -...

Question

$\frac{s e c 8 A - 1}{s e c 4 A - 1}$ is equal to

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

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

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$\frac{c o t 8 A}{c o t 2 A}$

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none of these

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Solution

The correct option is B
$\frac{t a n 8 A}{t a n 2 A}$

We have

$\frac{s e c 8 A - 1}{s e c 4 A - 1} = \frac{\frac{1}{c o s 8 A} - 1}{\frac{1}{c o s 4 A} - 1} = \frac{c o s 4 A}{c o s 8 A} \times \frac{1 - c o s 8 A}{1 - c o s 4 A} = \frac{c o s 4 A}{c o s 8 A} \times \frac{2 s i n^{2} 4 A}{2 s i n^{2} 2 A} (2 s i n^{2} θ = 1 - c o s 2 θ) = \frac{(2 c o s 4 A s i n 4 A) s i n 4 A}{2 \times c o s 8 A s i n^{2} 2 A}$

$= \frac{s i n 8 A s i n 4 A}{c o s 8 A \times 2 s i n 2 A \times s i n 2 A} = t a n 8 A \times \frac{2 s i n 2 A \times c o s 2 A}{2 s i n 2 A \times s i n 2 A} = t a n 8 A \times c o t 2 A = \frac{t a n 8 A}{t a n 2 A}$

Suggest Corrections

Similar questions

Q. Solve:

\frac{sec 8 A - 1}{sec 4 A - 1} - \frac{tan 8 A}{tan 4 A}

Q. Prove that

\frac{sec 8 A - 1}{sec 4 A - 1} = \frac{tan 8 A}{tan 2 A}

Q. (sec⁴ A − sec² A) = ?
(a) tan⁴ A − tan² A
(b) tan² A + tan⁴ A
(c) tan² A − tan⁴ A
(d) None of these

\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}

is equal to
(a) 0
(b) 1/2
(c) − 1
(d) none of these

{(\frac{- 3}{2})}^{- 1}

is equal to
(a)

\frac{2}{3}

(b)

- \frac{2}{3}

(c)

\frac{3}{2}

(d) none of these

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sec 8A−1sec 4A−1 is equal to

$\frac{s e c 8 A - 1}{s e c 4 A - 1}$ is equal to