Question

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Answer 1

Step-by-step explanation:

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Answer 2

Question:

Prove that:

√[ ( sec A + tan A ) / ( sec A - tan A ) ] = ( 1 + sin A ) / cos A

Answer:

√[ ( sec A + tan A ) / ( sec A - tan A ) ] = ( 1 + sin A ) / cos A

Step-by-step-explanation:

The given trigonometric equation is

√[ ( sec A + tan A ) / ( sec A - tan A ) ] = ( 1 + sin A ) / cos A

We have to prove this equation.

Now,

√[ ( sec A + tan A ) / ( sec A - tan A ) ] = ( 1 + sin A ) / cos A

∴ RHS = ( 1 + sin A ) / cos

⇒ RHS = ( 1 / cos A ) + ( sin A / cos A )

⇒ RHS = sec A + tan A

⇒ RHS = ( sec A + tan A ) / 1

Taking square and square root on both numerator and denominator, we get,

⇒ RHS = √[ ( sec A + tan A )² / 1² ]

⇒ RHS = √[ ( sec A + tan A )² / 1 ]

We know that,

1 + tan² A = sec² A

∴ sec² A - tan² A = 1

⇒ RHS = √[ ( sec A + tan A )² / ( sec² A - tan² A ) ]

We know that

a² - b² = ( a + b ) ( a - b )

⇒ RHS = √{ ( sec A + tan A )² / [ ( sec A + tan A ) ( sec A - tan A ) ] }

⇒ RHS = √{ ( sec A + tan A ) ( sec A + tan A ) / [ ( sec A + tan A ) ( sec A - tan A ) ] }

⇒ RHS = √[ ( sec A + tan A ) / ( sec A - tan A ) ]

LHS = √[ ( sec A + tan A ) / ( sec A - tan A ) ]

∴ LHS = RHS

Hence proved!