Prove the following identity, where the angle involved is acute angle for which the expression are defined.cosA+sinA−1cosA−sinA+1=cscA+cotA, Using the identity csc2A=1+cot2A.

LHS=cosA+sinA−1cosA−sinA+1=cotA+1−cscAcosA+sinA−1sinA=cotA−cscA+1(cotA+cscA)−(csc2−cot2A)[∵csc2A−cot2A=1]=cotA−cscA+1(cotA+cscA)−[(cscA+cotA)(cscA−cotA)]=(cotA−cscA+1)(cscA+cotA)(1−cscA+cotA)=cscA+cotA=RHS.

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Prove the following identity, where the angle involved is acut

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Prove the following identity, where the angle involved is acute angle for which the expression are defined.
$\frac{cos A - sin A + 1}{cos A + sin A - 1} = csc A + cot A$ , Using the identity $csc^{2} A = 1 + cot^{2} A$ .

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Prove the following identity, where the angle involved is acute angle for which the expression are defined.

\frac{cos A - sin A + 1}{cos A + sin A - 1} = csc A + cot A

, Using the identity

csc^{2} A = 1 + cot^{2} A

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Text solutionVerified

L H S = \frac{cos A - sin A + 1}{cos A + sin A - 1} = \frac{sin A}{\frac{cos A + sin A - 1}{cot A + 1 - csc A}}

= \frac{( cot A + csc A ) - ( csc ^{2} - cot ^{2} A )}{cot A - csc A + 1} [∵ csc^{2} A - cot^{2} A = 1]

= \frac{( cot A + csc A ) - [( csc A + cot A ) ( csc A - cot A )]}{cot A - csc A + 1}

= \frac{( csc A + cot A ) ( 1 - csc A + cot A )}{( cot A - csc A + 1 )}

= csc A + cot A = R H S

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Question Text	Prove the following identity, where the angle involved is acute angle for which the expression are defined. $\frac{cos A - sin A + 1}{cos A + sin A - 1} = csc A + cot A$ , Using the identity $csc^{2} A = 1 + cot^{2} A$ .
Answer Type	Text solution:1
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Prove the following identity, where the angle involved is acute angle for which the expression are defined.cosA+sinA−1cosA−sinA+1​=cscA+cotA, Using the identity csc2A=1+cot2A.

Text solutionVerified

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Prove the following identity, where the angle involved is acute angle for which the expression are defined.
$\frac{cos A - sin A + 1}{cos A + sin A - 1} = csc A + cot A$ , Using the identity $csc^{2} A = 1 + cot^{2} A$ .