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Integration Using Substitution

Is LHS=RHS ? ...

Question

Is LHS=RHS ?
$\frac{sec θ + 1 - tan θ}{sec θ + 1 + tan θ} + \frac{tan θ + sec θ - 1}{tan θ - sec θ + 1} = 2 sec θ$

Say true or false.

A

True

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B

False

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C

Ambiguous

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D

Data insufficient

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Solution

The correct option is A True
$= \frac{t a n A + s e c A - 1}{t a n A - s e c A + 1}$
$= \frac{s i n A + 1 - c o s A}{s i n A - 1 + c o s A}$
$= \frac{s i n A + (1 - c o s A)}{s i n - (1 - c o s A)}$
$= \frac{(s i n A + 1 - c o s A)^{2}}{s i n^{2} A - (1 - c o s A)^{2}}$
$= \frac{s i n^{2} A + c o s^{2} + 1 - 2 s i n A c o s A - 2 c o s A + 2 s i n A}{(1 - c o s^{2} A) - (1 - c o s A)^{2}}$
$= \frac{2 - 2 s i n A c o s A - 2 c o s A + 2 s i n A}{(1 - c o s A) (1 + c o s A - 1 + c o s A)}$
$= \frac{2 (1 - c o s A) + 2 s i n A (1 - c o s A)}{(1 - c o s A) (2 c o s A)}$
$= s e c A + t a n A$ ...... $(i)$
$A n d$

$\frac{s e c A + 1 - t a n A}{s e c A + 1 + t a n A}$
$= \frac{1 + c o s A - s i n A}{1 + c o s A + s i n A}$
$= \frac{(1 + c o s A - s i n A)^{2}}{(1 + c o s A)^{2} - s i n^{2} A}$
$= \frac{1 + c o s^{2} A + s i n^{2} A - 2 c o s A s i n A - 2 s i n A + 2 c o s A}{(1 + c o s A)^{2} - (1 - c o s A) (1 + c o s A)}$
$= \frac{2 - 2 c o s A s i n A - 2 s i n A + 2 c o s A}{(1 + c o s A) (1 + c o s A - 1 + c o s A)}$
$= \frac{2 (1 + c o s A s i n A - s i n A + c o s A)}{2 c o s A (1 + c o s A)}$
$= \frac{1 + c o s A s i n A - s i n A + c o s A}{c o s A (1 + c o s A)}$
$= \frac{(1 + c o s A) - s i n A (1 + c o s A)}{c o s A (1 + c o s A)}$
$= s e c A - t a n A$ ...(ii)
Adding $(i) a n d (i i)$ , we get
$2 s e c A$

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Similar questions

State whether true or false.

\frac{tan θ + sec θ - 1}{tan θ - sec θ + 1} = sec θ + tan θ .

\frac{sec θ + 1 - tan θ}{sec θ + 1 + tan θ} + \frac{tan θ + sec θ - 1}{tan θ - sec θ + 1} = 2 sec θ

\frac{tan θ}{(sec θ - 1)} + \frac{tan θ}{(sec θ + 1)} = 2 csc θ

\frac{tan θ}{sec θ - 1} = \frac{tan θ + sec θ + 1}{tan θ + sec θ - 1}

\frac{cot θ}{1 + tan θ} = \frac{cot θ - 1}{2 - {sec}^{2} θ}

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