Question

find quadratic polynomial p(x) if sum of product of zeroes are -2 and 2 out of respectively​

Answer 1

Given:

$\\ \red \bigstar \tt \: \frac{1 + sin \theta - cos \theta}{1 + sin \theta + cos \theta} = \frac{tan \theta}{2}$

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To Prove:

$\\ \purple\bigstar \tt \: \frac{1 + sin \theta - cos \theta}{1 + sin \theta + cos \theta} = \frac{tan \theta}{2}$

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Formulas:

$\underline{\bf{\bullet \: Using \: the \: trigonometry \: identities \: we \: can \: write, }}$

$\\ \tt \: (1) \: sin \theta = 2sin\frac{ \theta}{2} cos \frac{ \theta}{2}$

$\\ \tt \: (2) \: cos \theta = 2 {cos}^{2} \frac{ \theta}{2} - 1$

$\\ \tt \: (3) \:cos \theta = 1 - 2 {sin}^{2} \frac{ \theta}{2}$

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Proof:

Taking LHS.

$\\ \implies \sf \: lhs \: = \frac{1 + sin \theta - cos \theta}{1 + sin \theta + cos \theta}$

Applying these identities :

$\\ \implies \sf lhs = \frac{1 + sin \theta - cos \theta}{1 + sin \theta + cos \theta} \\ \\ \\ \implies \sf \: lhs = \frac{1 + 2sin \frac{ \theta}{2}cos \frac{ \theta}{2} - 1 + 2 {sin}^{2} \frac{ \theta}{2} }{1 + 2sin \frac{ \theta}{2}cos \frac{ \theta}{2} + 2 {cos}^{2} \frac{ \theta}{2} - 1} \\ \\ \\ \implies \sf \: lhs = \frac{2sin \frac{ \theta}{2} cos \frac{ \theta}{2} + 2 {sin}^{2} \frac{ \theta}{2} }{2sin \frac{ \theta}{2}cos \frac{ \theta}{2} + 2 {cos}^{2} \frac{ \theta}{2} } \\ \\ \\ \implies \sf \: lhs = \frac{2sin \frac{ \theta}{2} cos \frac{ \theta}{2} + sin \frac{ \theta}{2} }{2cos \frac{ \theta}{2}sin \frac{ \theta}{2} + cos \frac{ \theta}{2} }$

Cancelling the like terms :

$\\ \implies \sf \: lhs = \frac{sin \frac{ \theta}{2} }{cos \frac{ \theta}{2} } \\ \\ \implies \sf \: lhs = tan \frac{ \theta}{2} \\ \\ \implies \sf \: lhs = rhs$

$\red \bigstar \boxed{ \sf{ \therefore \: \frac{1 + sin \theta - cos \theta}{1 + sin \theta + cos \theta} = tan \frac{ \theta}{2} }}$

•Hence proved!