Question

prove the problems given below

Answer 1

Solution ( i ) :-

We need to prove ,

• sin²α - cos²β = sin²β - cos²α

Simplifying ,

⇒ sin²α - cos²β = sin²β - cos²α

⇒ sin²α + cos²α = sin²β + cos²β

⇒ 1 = 1 [ ∵ sin²A + cos²A = 1 ]

• LHS = RHS

Hence , proved .

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Solution ( ii ) :-

We need to prove ,

• tanθ ( √ 1 - sin²θ ) = sinθ

Taking LHS and substituting ,

• √ 1 - sin²θ = cosθ

⇒ tanθ ( cosθ )

Substituting ,

• tanθ = sinθ/cosθ

⇒ [ sinθ/cosθ ] [ cosθ ]

⇒ sinθ

Now , comparing with RHS

⇒ sinθ = sinθ

• LHS = RHS

Hence , proved .

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Solution ( iii ) :-

We need to prove ,

• ( 1 - cos²θ ) cosec²θ = 1

Taking LHS & simplifying ,

⇒ ( 1 - cos²θ ) cosec²θ

⇒ sin²θ × cosec²θ [ ∵ 1 - cos²A = sin²A ]

⇒ sin²θ × ( 1/sin²θ ) [ ∵ cosec²A = 1/sin²A ]

⇒ 1

Now , comparing with RHS

⇒ 1 = 1

• LHS = RHS

Hence , proved .

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Solution ( iv ) :-

We need to prove ,

• ( 1 + tanA )² + ( 1 - tanA )² = 2sec²A

Taking LHS and simplifying using

• ( a + b )² = a² + b² + 2ab

• ( a - b )² = a² + b² - 2ab

⇒ 1² + tan²A + 2 ( 1 ) ( tanA ) + 1² + tan²A - 2 ( 1 ) ( tanA )

⇒ 1 + tan²A + 2tanA + 1 + tan²A - 2tanA

⇒ sec²A + sec²A [ ∵ 1 + tan²A = sec²A ]

⇒ 2sec²A

Now, comparing with RHS

⇒ 2sec²A = 2sec²A

• LHS = RHS

Hence , proved

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Solution ( v ) :-

We need to prove ,

• cos²β + ( 1/1 + cot²β ) = 1

Taking LHS & simplifying ,

⇒ cos²β + ( 1/cosec²β ) [ ∵ 1 + cot²A = csc²A ]

⇒ cos²β + sin²β [ ∵ 1/cosec²β = sin²β ]

⇒ 1 [ ∵ sin²A + cos²A = 1 ]

Now , comparing with RHS

⇒ 1 = 1

• LHS = RHS

Hence , proved .