Question

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Answer 1

Answer:

p^2 = cosecthita-sinthita , q^2= secthita-cosecthita

Step-by-step explanation:

cosecthita^2-sinthita^2×secthita^2-cosecthita^2(cosecthita^2-sinthita+secthita-cosecthita^2)=1

= -sinthita^2×secthita^2(-sinthita^2+secthita^2) {+ + - = 0}

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Answer 2

Solution: Given: cosecβ-sinβ = p and, secβ-cosβ = q.

⇒ p² = (cosecβ-sinβ)² = cosec²β+sin²β-2cosec.βsinβ = cosec²β+sin²β-2 = cosec²β-1+sin²β-1

⇒ p² = cot²β-cos²β

And q² = (secβ-cosβ)² = sec²β+cos²β-2secβ.cosβ = sec²β+cos²β-2 = sec²β-1+cos²β-1

⇒ q² = tan²β-sin²β

To prove: p²q²(p²+q²+3) = 1

Taking L.H.S.,

= p²q²(p²+q²+3)

= (cot²β-cos²β)(tan²β-sin²β)(cot²β-cos²β+tan²β-sin²β+3)

= [cot²β.tan²β-cos²β.tan²β-cot²β.sin²β+cos²β.sin²β][cot²β+tan²β-(cos²β+sin²β)+3]

= [1-sin²β-cos²β+cos²β.sin²β][cot²β+tan²β-1+3]

= [1-(sin²β+cos²β)+cos²β.sin²β][cot²β+tan²β+2]

= [1-1+cos²β.sin²β][cot²β+tan²β+2]

= cos²β.sin²β.[cot²β+tan²β+2]

= cos²β.sin²β.cot²β + cos²β.sin²β.tan²β + 2cos²β.sin²β

= cos⁴β + sin⁴β + 2cos²β.sin²β

= (cos²β+sin²β)²

= 1²

= 1 = R.H.S.

Hence Proved

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