Question

Trigeonometric Identities
Exercise 13 A
Question number 33
$\sf{ \{ \frac{1}{(sec^2 \theta - cos^2 \theta )}} + \frac{1}{(cosec^2 \theta - sin^2\theta )} \} ( sin^2\theta \times cos^2\theta)} = \frac{1 - sin^2 \theta \times cos^2\theta}{2 + sin^2\theta \times cos^2\theta} }$


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Answer 1

Hope so that it was useful

Answer 2

Answer:

LHS = RHS

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Step-by-step explanation:

Taking the LHS we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{1}{\big(sec^{2}\theta - cos^{2}\theta\big)} + \dfrac{1}{\big(cosec^{2}\theta - sin^{2}\theta\big)}\Bigg\}\times \Big(sin^{2}\theta \times cos^{2}\theta\Big)$

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All the terms in the RHS of the given equation are in terms of sinθ and cosθ, so let's try to express the LHS in terms of the same.

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We know that;

• sec²θ = 1/cos²θ

• cosec²θ = 1/sin²θ

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$\sf \dashrightarrow \Bigg\{\dfrac{1}{\bigg(\dfrac{1}{cos^{2}\theta} - cos^{2}\theta\bigg)} + \dfrac{1}{\bigg(\dfrac{1}{sin^{2}\theta} - sin^{2}\theta\bigg)}\Bigg\}\times \Big(sin^{2}\theta \times cos^{2}\theta\Big)$

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On taking LCM (in the denominator) we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{1}{\bigg(\dfrac{1 - cos^{4}\theta}{cos^{2}\theta}\bigg)} + \dfrac{1}{\bigg(\dfrac{1 - sin^{4}\theta}{sin^{2}\theta}\bigg)}\Bigg\}\times \Big(sin^{2}\theta \times cos^{2}\theta\Big)$

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{2}\theta}{1 - cos^{4}\theta} + \dfrac{sin^{2}\theta}{1 - sin^{4}\theta}\Bigg\}\times \Big(sin^{2}\theta \times cos^{2}\theta\Big)$

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On applying a² - b² = (a + b) (a - b) in the denominator we get;

[For the first fraction, a = 1, b = cos²θ]

[For the second fraction, a = 1, b = sin²θ]

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{2}\theta}{(1 + cos^{2}\theta)(1 - cos^{2}\theta)} + \dfrac{sin^{2}\theta}{(1 + sin^{2}\theta)(1 - sin^{2}\theta)}\Bigg\}\times \Big(sin^{2}\theta \times cos^{2}\theta\Big)$

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We know that;

• 1 - cos²θ = sin²θ

• 1 - sin²θ = cos²θ

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{2}\theta}{(1 + cos^{2}\theta) \times sin^{2}\theta} + \dfrac{sin^{2}\theta}{(1 + sin^{2}\theta) \times cos^{2}\theta}\Bigg\}\times \Big(sin^{2}\theta \times cos^{2}\theta\Big)$

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On taking LCM we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta(1 + sin^{2}\theta) + sin^{4}\theta(1 + cos^{2}\theta)}{(1 + cos^{2}\theta)(1 + sin^{2}\theta)(sin^{2}\theta \times cos^{2}\theta)}\Bigg\}\times \Big(sin^{2}\theta \times cos^{2}\theta\Big)$

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On cancelling (sin²θ × cos²θ) we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta(1 + sin^{2}\theta) + sin^{4}\theta(1 + cos^{2}\theta)}{(1 + cos^{2}\theta)(1 + sin^{2}\theta)}\Bigg\}$

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On cross-multiplying we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta + cos^{4}\theta \ sin^{2}\theta + sin^{4}\theta + sin^{4}\theta \ cos^{2}\theta}{(1 + cos^{2}\theta)(1 + sin^{2}\theta)}\Bigg\}$

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta + sin^{4}\theta + cos^{4}\theta \ sin^{2}\theta + sin^{4}\theta \ cos^{2}\theta}{1 + sin^{2}\theta + cos^{2}\theta + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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In the numerator, cos⁴θ sin²θ + sin⁴θ cos²θ in the numerator can be re-written as sin²θcos²θ(cos²θ + sin²θ)

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta + sin^{4}\theta + (sin^{2}\theta \ cos^{2}\theta)(cos^{2}\theta + sin^{2}\theta)}{1 + sin^{2}\theta + cos^{2}\theta + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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On substituting sin²θ + cos²θ = 1 we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta + sin^{4}\theta + (sin^{2}\theta \ cos^{2}\theta)(1)}{1 + 1 + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta + sin^{4}\theta + sin^{2}\theta \ cos^{2}\theta}{2 + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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On adding and subtracting sin²θcos²θ in the numerator we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta + sin^{4}\theta + sin^{2}\theta \ cos^{2}\theta + sin^{2}\theta \ cos^{2}\theta - sin^{2}\theta \ cos^{2}\theta}{2 + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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$\sf \dashrightarrow \Bigg\{\dfrac{cos^{4}\theta + sin^{4}\theta + 2sin^{2}\theta \ cos^{2}\theta - sin^{2}\theta \ cos^{2}\theta}{2 + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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The numerator is in the form of (a + b)² = a² + b² + 2ab, where a = sin²θ and b = cos²θ.

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$\sf \dashrightarrow \Bigg\{\dfrac{(cos^{2}\theta + sin^{2}\theta)^{2} - sin^{2}\theta \ cos^{2}\theta}{2 + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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On substituting sin²θ + cos²θ = 1 we get;

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$\sf \dashrightarrow \Bigg\{\dfrac{(1)^{2} - sin^{2}\theta \ cos^{2}\theta}{2 + sin^{2}\theta cos^{2}\theta}\Bigg\}$

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$\sf \dashrightarrow \Bigg\{\dfrac{1 - sin^{2}\theta \ cos^{2}\theta}{2 + sin^{2}\theta \ cos^{2}\theta}\Bigg\}$

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LHS = RHS

Hence proved.