Question

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9. (1 – tan B)² + (1 - cot B)2 = (sec B - cosec B)?​

Answer 1

Correct Question :

Solve this question :

• (1 - tan B)² + (1 - cot B)² = (sec B - cosecB)²

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Explanation :

To Prove :

• (1 - tan B)² + (1 - cot B)² = (sec B - cosecB)²

Proof :

LHS = (1 - tan B)² + (1 - cot B)²

By using identinty :

• (a - b)² = a² + b² - 2ab

$\sf : \implies \underline{\bf LHS} = [\pink{(1)^{2} + (tan B)^{2} - 2 \times 1 \times tanB}] + [\pink{(1)^{2} + (cot B)^{2} - 2 \times 1 \times cotB}]$

$\sf : \implies \underline{\bf LHS} = [1 + tan^{2} B - 2tanB] + [1 + cot^{2}B - 2cotB]$

Now, we know that :

• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

$\sf : \implies \underline{\bf LHS} = [\pink{sec^{2} B} - 2tanB] + [\pink{cosec^{2}B} - 2cotB]$

$\sf : \implies \underline{\bf LHS} = sec^{2} B - 2tanB + cosec^{2}B - 2cotB$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2tanB - 2cotB$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2(tanB + cotB)$

Now, we know that :

• $\sf tan \theta = \dfrac{sin \theta}{cos \theta}$
• $\sf cot \theta = \dfrac{cos \theta}{sin \theta}$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 \Bigg( \pink{\dfrac{sin B}{cos B}} + \pink{\dfrac{cos B}{sin B}} \Bigg)$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 \Bigg( \dfrac{sin B \times sin B}{cos B \times sin B} + \dfrac{cos B \times cos B}{sin B \times cos B} \Bigg)$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 \Bigg( \dfrac{sin^{2} B}{sin B cos B} + \dfrac{cos^{2} B}{sin Bcos B} \Bigg)$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 \Bigg(\dfrac{sin ^{2} B + cos^{2} B}{sin Bcos B} \Bigg)$

Now, we know that :

• sin²θ + cos²θ = 1

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 \Bigg(\dfrac{\pink{1}}{sin Bcos B} \Bigg)$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 \Bigg(\dfrac{1}{sin B} \times \dfrac{1}{cos B} \Bigg)$

Now, we know that :

• $\sf \dfrac{1}{sin \theta} = cosec \theta$
• $\sf \dfrac{1}{cos \theta} = sec \theta$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 (\pink{cosec \theta} \times \pink{sec \theta})$

$\sf : \implies \underline{\bf LHS} = sec^{2} B + cosec^{2}B - 2 sec \theta \cosec \theta$

Now, by using identinty :

• a² + b² - 2ab = (a - b)²

$\sf : \implies \underline{\bf LHS} = (secB + cosecB)^{2}$

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RHS = (sec B - cosecB)²

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As, LHS = RHS,

Hence, verified.