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Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

i Evaluate ...

Question

(i) Evaluate $(1 + tan θ + sec θ) (1 + cot θ - c o sec θ)$
(ii) Prove that $\frac{tan A - sin A}{tan A + sin A} = \frac{sec A - 1}{sec A + 1}$

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Solution

(ii)

$\frac{tan A - sin A}{tan A + sin A} = \frac{sec A - 1}{sec A + 1}$

Taking LHS

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

$L H S = R H S$

Hence proved.

(i)

$(1 + tan A + sec A) (1 + cot A - c o s e c A)$
$(1 + \frac{sin A}{c o s A} + \frac{1}{c o s A}) (1 + \frac{c o s A}{s i n A} - \frac{1}{s i n A})$
$= (\frac{c o s A + s i n A + 1}{c o s A}) (\frac{c o s A + s i n A - 1}{s i n A})$
$= \frac{{(c o s A + s i n A)}^{2} - 1}{s i n A c o s A}$
$= \frac{{sin}^{2} A + {cos}^{2} A + 2 s i n A c o s A - 1}{s i n A c o s A}$
$= \frac{1 + 2 s i n A c o s A - 1}{s i n A c o s A}$
$= 2$

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Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

Standard XII Mathematics

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