Which of the following options is equal to the given expresssion?

$\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ $\times$ $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$

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Solution

The correct option is A

cot( $90^{\circ} - θ$ ) = $t a n θ$

sin( $90^{\circ} - θ$ ) = $c o s θ$

LHS = $\frac{t a n θ}{c o s e c^{2} θ}$ $\times$ $\frac{s e c θ c o t^{3} θ}{c o s^{2} θ}$

= $t a n θ$ $\times$ $\frac{s i n^{2} θ}{c o s^{2} θ}$ $\times$ $s e c θ$ $\times$ $c o t^{3} θ$

= $t a n θ$ $\times$ $t a n^{2} θ$ $\times$ $s e c θ$ $\times$ $c o t^{3} θ$

= $t a n^{3} θ$ $\times$ $c o t^{3} θ$ $\times$ $s e c θ$

= $s e c θ$

= $\sqrt{1 + t a n^{2} θ}$

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Similar questions

Which of the following options is equal to the given expresssion?

$\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ $\times$ $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$

Show that: $\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ . $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$ = $\sqrt{t a n^{2} θ + 1}$

Which of the following options is equal to $\frac{sin θ + 1 - cos θ}{cos θ - 1 + sin θ} ?$

Which of the following options is equal to $\frac{c o t θ + c o s e c θ - 1}{c o t θ - c o s e c θ + 1}$

c o t^{2} - c o s e c^{2} θ =

(0 < θ < 90)

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