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Answer 1

Answer:

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Answer 2

Question:-

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution:-

(i) (cosecθ - cotθ)² = $\sf{\dfrac{1 + cos\theta}{1 + cos\theta}}$

⟶ Taking LHS,

We know,

• cosecθ = $\sf{\dfrac{1}{sin\theta}}$

• cotθ = $\sf{\dfrac{cos\theta}{sin\theta}}$

Hence,

(cosecθ - cotθ)²

= $\sf{\bigg(\dfrac{1}{sin\theta} - \dfrac{cos\theta}{sin\theta}\bigg)^2}$

= $\sf{\bigg(\dfrac{1 - cos\theta}{sin\theta}\bigg)^2}$

= $\sf{\dfrac{(1 - cos\theta)^2}{sin^2\theta}}$

= $\sf{\dfrac{(1 - cos\theta)(1 - cos\theta)}{1 - cos^2\theta}}$

= $\sf{\dfrac{(1 - cos\theta)(1 - cos\theta)}{(1 - cos\theta)(1 + cos\theta)}}$

= $\sf{\dfrac{1 - cos\theta}{1 + cos\theta}}$

Taking RHS,

= $\sf{\dfrac{1 - cos\theta}{1 + cos\theta}}$

Hence,

LHS = RHS (Proved)

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(ii) $\sf{\dfrac{cosA}{1 + sinA} + \dfrac{1 + sinA}{cosA} = 2 secA}$

⟶ Taking LHS,

$\sf{\dfrac{cosA}{1 + sinA} + \dfrac{1 + sinA}{cosA}}$

Taking LCM,

= $\sf{\dfrac{cos^2A + (1 + sinA)^2}{cosA(1 + sinA)}}$

= $\sf{\dfrac{cos^2A + 1 + 2sinA + sin^2A}{cosA(1 + sinA)}}$

= $\sf{\dfrac{cos^2A + sin^2A + 1 + 2sinA}{cosA(1 + sinA)}}$

= $\sf{\dfrac{1 + 1 + 2sinA}{cosA(1 + sinA)}}$

= $\sf{\dfrac{2 + 2sinA}{cosA(1 + sinA)}}$

= $\sf{\dfrac{2(1 + sinA)}{cosA(1 + sinA)}}$

= $\sf{\dfrac{2}{cosA}}$

We know,

$\sf{\dfrac{1}{cosA} = secA}$

Hence,

= $\sf{2 \times\dfrac{1}{cosA}}$

= 2secA

Taking RHS,

2secA

Hence,

LHS = RHS (Proved)

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(iii) $\sf{\dfrac{tan\theta}{1 - cot\theta} + \dfrac{cot\theta}{1 - tan\theta} = 1 + sec\theta cosec\theta}$

⟶ Refer to the first two attachments for solution.

Identities used:-

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

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

a³ - b³ = (a - b)(a² + ab + b²)

$\sf{\dfrac{1}{sin\theta} = cosec\theta}$

$\sf{\dfrac{1}{cos\theta} = sec\theta}$

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(iv) $\sf{\dfrac{1 + secA}{secA} = \dfrac{sin^2A}{1 - cosA}}$

⟶ Taking LHS,

We know,

$\sf{secA = \dfrac{1}{cosA}}$

Hence,

$\sf{\dfrac{1 + \dfrac{1}{cosA}}{\dfrac{1}{cosA}}}$

= $\sf{\dfrac{cosA + 1}{cosA} \times cosA}$

= $\sf{cosA + 1}$

Taking RHS,

We know,

sin²A = 1 - cos²A

$\sf{\dfrac{sin^2A}{1 - cosA}}$

= $\sf{\dfrac{1 - cos^2A}{1 - cosA}}$

= $\sf{\dfrac{(1 + cosA)(1 - cosA)}{1 - cosA)}}$

= $\sf{cosA + 1}$

Hence,

LHS = RHS (Proved)

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(v) $\sf{\dfrac{cosA - sinA + 1}{cosA + sinA - 1} = cosecA + cotA}$

⟶ Refer to the next two attachments for solution:-

Identity used:-

a² - b² = (a + b)(a - b)

And Hence,

LHS = RHS (Proved)

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