Question

whole root cosectheta-1/cosectheta+1 + whole root cosectheta+1/ cosectheta-1 = 2sec theta​

Answer 1

Answer:

your answer is given in the above photo

Answer 2

To prove :-

$\rm { \dashrightarrow \sqrt{ \dfrac{ cosec \theta - 1}{ cosec \theta + 1} } + \sqrt{ \dfrac{ cosec \theta + 1}{ cosec \theta - 1} } = 2 \sec \theta }$

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Solution :-

» Consider LHS

$\rm { \dashrightarrow \sqrt{ \dfrac{ cosec \theta - 1}{ cosec \theta + 1} } + \sqrt{ \dfrac{ cosec \theta + 1}{ cosec \theta - 1} } }$

» Rationalise dn. of fractions

$\rm { \dashrightarrow \sqrt{ \dfrac{ cosec \theta - 1}{ cosec \theta + 1} \times\dfrac{ cosec \theta - 1}{ cosec \theta - 1} } + \sqrt{ \dfrac{ cosec \theta + 1}{ cosec \theta - 1} \times \dfrac{ cosec \theta + 1} { cosec \theta + 1}} }$

$\rm { \dashrightarrow \sqrt{ \dfrac{ (cosec \theta - 1) ^{2} }{ ( cosec \theta + 1)(cosec \theta - 1)} } + \sqrt{ \dfrac{ ( cosec \theta + 1)^{2} }{ ( cosec \theta + 1)(cosec \theta - 1)}} }$

» (cosec A + 1)(cosec A - 1)= cosec²A -1

$\rm { \dashrightarrow \sqrt{ \dfrac{ (cosec \theta - 1) ^{2} }{ (cosec ^{2} \theta - 1)} } + \sqrt{ \dfrac{ ( cosec \theta + 1)^{2} }{ (cosec ^{2} \theta - 1)}} }$

» cosec²A - 1 = cot²A

$\rm { \dashrightarrow \sqrt{ \dfrac{ (cosec \theta - 1) ^{2} }{ (cot ^{2} \theta )} } + \sqrt{ \dfrac{ ( cosec \theta + 1)^{2} }{ (cot ^{2} \theta)}} }$

$\rm { \dashrightarrow \sqrt{ \bigg( \dfrac{ cosec \theta - 1 }{ \cot \theta } \bigg) ^{2} } + \sqrt{ \bigg(\dfrac{ cosec \theta + 1}{ \cot \theta} \bigg) ^{2} } }$

$\rm { \dashrightarrow \dfrac{ cosec \theta - 1 }{ \cot \theta } } + \dfrac{ cosec \theta + 1}{ \cot \theta}$

$\rm { \dashrightarrow \dfrac{ cosec \theta - 1 +cosec \theta + 1 }{ \cot \theta } }$

$\rm { \dashrightarrow \dfrac{ 2cosec \theta }{ \cot \theta } }$

» cot A = cosec A/sec A

$\rm { \dashrightarrow \bigg(\dfrac{ 2cosec \theta }{ \frac{ cosec \theta}{ \sec \theta} } \bigg) }$

$\rm { \dashrightarrow \bigg(\dfrac{ 2cosec \theta \times \sec \theta}{ cosec \theta} \bigg) }$

$\rm { \dashrightarrow \bigg(\dfrac{ 2 \times \sec \theta}{1} \bigg) }$

${ \dashrightarrow \boxed{ \green{2 \sec \theta}}}$

» LHS = RHS

Hence proved