Prove that1-sec -tan -1-cos - - 1-cos -1-sec -tan - -4 MARKS-

Concept: 2 Marks Application: 2 Marks We have LHS 1/sec θ tan θ 1/cos θ =1/(sec θ tan θ)×(sec θ+tan θ)/(sec θ+tan θ) sec θ =(sec θ+tan θ)/(sec^2 θ tan^2 θ) ...

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Byju's Answer

Standard XII

Mathematics

Range of Trigonometric Expressions

Prove that1/s...

Question

Prove that

$\frac{1}{(s e c θ - t a n θ)} - \frac{1}{c o s θ} = \frac{1}{c o s θ} - \frac{1}{(s e c θ + t a n θ)}$ [4 MARKS]

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Solution

Concept: 2 Marks
Application: 2 Marks

We have

LHS
$\frac{1}{s e c θ - t a n θ} - \frac{1}{c o s θ}$

$= \frac{1}{(s e c θ - t a n θ)} \times \frac{(s e c θ + t a n θ)}{(s e c θ + t a n θ)} - s e c θ$

$= \frac{(s e c θ + t a n θ)}{(s e c^{2} θ - t a n^{2} θ)} - s e c θ$

$= (s e c θ + t a n θ) - s e c θ$ $[∵ s e c^{2} θ - t a n^{2} θ = 1]$

$= t a n θ$

RHS
$\frac{1}{c o s θ} - \frac{1}{(s e c θ + t a n θ)}$

$= s e c θ - \frac{1}{(s e c θ + t a n θ)} \times \frac{(s e c θ - t a n θ)}{(s e c θ - t a n θ)}$

$= s e c θ - \frac{(s e c θ - t a n θ)}{(s e c^{2} θ - t a n^{2} θ)}$

$= s e c θ - (s e c θ - t a n θ)$ $[∵ s e c^{2} θ - t a n^{2} θ = 1]$

$= t a n θ$

$∴$ LHS = RHS

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Prove that

$\frac{1}{(s e c θ - t a n θ)} - \frac{1}{c o s θ} = \frac{1}{c o s θ} - \frac{1}{(s e c θ + t a n θ)}$ [4 MARKS]