The correct option is C tan θ1/cos θ 1/(sec θ+tan θ) =sec 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) =sec θ (sec θ ...

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Byju's Answer

Standard X

Mathematics

Relation between Trigonometric Ratios

1/cosθ-1/θ+ta...

Question

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

s i n θ

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c o t θ

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c o s θ

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t a n θ

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Solution

The correct option is D $t a n θ$
$\frac{1}{c o s θ} - \frac{1}{(s e c θ + t a n θ)}$

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

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

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

$= t a n θ$

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Similar questions

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

Q. Solve:

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

Q. Prove that:

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

Q. Question 12
Prove that

\frac{1 + s e c θ - t a n θ}{1 + s e c θ + t a n θ} = \frac{1 - s i n θ}{c o s θ}

The expression $E = \frac{sec θ + tan θ - 1}{tan θ - sec θ + 1}$ can be simplified to -

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1cos θ−1(sec θ+tan θ)

The correct option is D tan θ1cos θ−1(sec θ+tan θ) =sec θ−1(sec θ+tan θ)×(sec θ−tan θ)(sec θ−tan θ) =sec θ−(sec θ−tan θ)(sec2 θ−tan2 θ) =sec θ−(sec θ−tan θ) [∵sec2 θ−tan2 θ=1] =tan θ

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