Which of the following options are equal to tan A- A-1-tan A- A-1 -

The correct option is D 2/sin A tan A/sec A−1+tan A/sec A+1 = tanA [1/sec A−1+1/sec A+1] = tanA [sec A+1+sec A−1/(sec A−1)(sec A+1)] = tanA [2sec A/(sec^2 A−1) ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Which of the ...

Question

Which of the following options are equal to $\frac{t a n A}{s e c A - 1} + \frac{t a n A}{s e c A + 1}$ ?

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

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$\frac{1}{s i n A}$

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$2 c o s e c A$

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Solution

The correct options are
B
$\frac{2}{s i n A}$

D
$2 c o s e c A$

$\frac{t a n A}{s e c A - 1} + \frac{t a n A}{s e c A + 1}$

= $t a n A [\frac{1}{s e c A - 1} + \frac{1}{s e c A + 1}]$

= $t a n A [\frac{s e c A + 1 + s e c A - 1}{(s e c A - 1) (s e c A + 1)}]$

= $t a n A [\frac{2 s e c A}{(s e c^{2} A - 1)}]$

= $t a n A [\frac{2 s e c A}{t a n^{2} A}]$

= $t a n A [\frac{2 s e c A}{t a n^{2} A}]$

= $\frac{2 s e c A}{t a n A}$

= $\frac{2}{c o s A \frac{s i n A}{c o s A}}$

= $\frac{2}{s i n A}$

= $2 c o s e c A$

Suggest Corrections

Which of the following options are equal to tanAsecA−1+tanAsecA+1 ?

The correct options are B 2sinA D 2cosecA tanAsecA−1+tanAsecA+1 = tanA[1secA−1+1secA+1] = tanA[secA+1+secA−1(secA−1)(secA+1)] = tanA[2secA(sec2A−1)] = tanA[2secAtan2A] = tanA[2secAtan2A] = 2secAtanA = 2cosAsinAcosA = 2sinA = 2cosecA

Which of the following options are equal to $\frac{t a n A}{s e c A - 1} + \frac{t a n A}{s e c A + 1}$ ?