JEE Advanced Trigonometric Functions PYQ – Trigonometric Ratios, Domain & Range

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Q 1 :

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as: [2013]

$\sin A \cos A + 1$
$\sec A c o s e c A + 1$
$\tan A + \cot A$
$\sec A + c o s e c A$

(2)

Given expression can be written as

$\frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A - \cos A} + \frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A - \sin A}$

$= \frac{1}{\sin A - \cos A} {\frac{\sin^{3} A - \cos^{3} A}{\cos A \sin A}}$

$∵$ $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$

$= \frac{\sin^{2} A + \sin A \cos A + \cos^{2} A}{\sin A \cos A} = 1 + \sec A cosec A$

Q 2 :

Given both $θ$ and $ϕ$ are acute angles and $\sin θ = \frac{1}{2}, \cos ϕ = \frac{1}{3},$ then the value of $θ + ϕ$ belongs to [2004]

$(\frac{π}{3}, \frac{π}{2}]$
$(\frac{π}{2}, \frac{2 π}{3})$
$(\frac{2 π}{3}, \frac{5 π}{6}]$
$(\frac{5 π}{6}, π]$

(2)

$Given: \sin θ = \frac{1}{2} and θ is acute angle$

$∴$ $θ = \frac{π}{6}$

$Also given, \cos ϕ = \frac{1}{3} and ϕ is acute angle.$

$∴$ $0 < \frac{1}{3} < \frac{1}{2}$

$\Rightarrow \cos \frac{π}{2} < \cos ϕ < \cos \frac{π}{3} or \frac{π}{3} < ϕ < \frac{π}{2}$

$∴$ $\frac{π}{3} + \frac{π}{6} < θ + ϕ < \frac{π}{2} + \frac{π}{6} or \frac{π}{2} < θ + ϕ < \frac{2 π}{3}$

$\Rightarrow θ + ϕ \in (\frac{π}{2}, \frac{2 π}{3})$

Topic Question Set

Q 1 :

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as: [2013]

Q 2 :

Given both $θ$ and $ϕ$ are acute angles and $\sin θ = \frac{1}{2}, \cos ϕ = \frac{1}{3},$ then the value of $θ + ϕ$ belongs to [2004]

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Topic Question Set

Q 1 : The expression tanA1-cotA+cotA1-tanA can be written as: [2013]

Q 2 : Given both θ and ϕ are acute angles and sinθ=12,cosϕ=13, then the value of θ+ϕ belongs to [2004]

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Q 1 :

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as: [2013]

Q 2 :

Given both $θ$ and $ϕ$ are acute angles and $\sin θ = \frac{1}{2}, \cos ϕ = \frac{1}{3},$ then the value of $θ + ϕ$ belongs to [2004]