First slide

Trigonometric Identities

Question

If $x = \cos e c^{2} θ : y = \sec^{2} θ, z = \frac{1}{1 - \sin^{2} θ \cos^{2} θ}$ they $x y z =$

Moderate

A

$x + y + z$
B

$x y + z$
C

$x + y - z$
D

$\frac{x + y}{z}$

Solution

$W e h a v e x = \cos e c^{2} θ, y = s e c^{2} θ a n d z = \frac{1}{1 - \sin^{2} θ \cos^{2} θ} N o w z = \frac{1}{1 - \sin^{2} θ \cos^{2} θ} = \frac{1}{1 - \frac{1}{x} \frac{1}{y}} = \frac{x y}{x y - 1} \Rightarrow z (x y - 1) = x y \Rightarrow x y z = x y + z \to (1)$
$A l s o x y = \cos e c^{2} θ s e c^{2} θ = \frac{1}{\sin^{2} θ \cos^{2} θ} = \frac{\sin^{2} θ + \cos^{2} θ}{\sin^{2} θ \cos^{2} θ}$
$= \frac{1}{\sin^{2} θ} + \frac{1}{\cos^{2} θ} = x + y \to (2) B y (1) a n d (2), w e h a v e x y z = x y + z = x + y + z$

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