Question

solve the above problem from the trigonometric chapter.....​

Answer 1

$\tt{Given \ :}$


$\tt{ {\dfrac{1 - cos \theta + sin \theta}{1 + sin \theta}} }$


$\tt{\implies {\dfrac{1 + sin \theta - cos \theta}{1 + sin \theta}} }$


$\tt{\implies {\dfrac{1 + sin \theta - cos \theta}{1 + sin \theta}} × {\dfrac{1 + sin \theta + cos \theta}{1 + sin \theta + cos \theta}} }$


$\tt{\implies {\dfrac{(1 + sin \theta - cos \theta) × (1 + sin \theta + cos \theta)}{(1 + sin \theta) × (1 + sin \theta + cos \theta}} }$


${\boxed{\tt{(a + b)(a - b) = a^2 - b^2}}}$


$\tt{Here, \ a = 1 + sin \theta, \ b = cos \theta}$


$\tt{\implies {\dfrac{(1 + sin \theta)^2 - (cos \theta)^2}{(1 + sin \theta)^2 + cos \theta ( 1 + sin \theta)}} }$


${\boxed{\tt{(a + b)^2 = a^2 + b^2 + 2ab}}}$


$\tt{\implies {\dfrac{1 + sin^2 \theta + 2sin \theta - cos^2 \theta}{1 + sin^2 \theta + 2sin \theta + cos \theta + cos \theta . sin \theta}} }$


${\boxed{\tt{1 = sin^2 \theta + cos^2 \theta}}}$


$\tt{\implies {\dfrac{sin^2 \theta + cos^2 \theta + sin^2 \theta + 2sin \theta - cos^2 \theta}{1 + sin^2 \theta + 2sin \theta + cos \theta + cos \theta . sin \theta}} }$


$\tt{\implies {\dfrac{sin^2 \theta + {\cancel{cos^2 \theta}} + sin^2 \theta + 2sin \theta - {\cancel{cos^2 \theta}} }{1 + sin^2 \theta + 2sin \theta + cos \theta + cos \theta . sin \theta}} }$


$\tt{\implies {\dfrac{2sin^2 \theta + 2sin \theta}{(1 + sin \theta)^2 + cos \theta ( 1 + sin \theta)}} }$


$\tt{\implies {\dfrac{2sin \theta (sin \theta + 1)}{(1 + sin \theta)[(1 + sin \theta) + cos \theta]}}}$


$\tt{\implies {\dfrac{2sin \theta ({\cancel{sin \theta + 1}})}{({\cancel{1 + sin \theta}})[(1 + sin \theta) + cos \theta]}}}$


$\tt{\implies {\dfrac{2sin \theta}{1 + sin \theta + cos \theta}}}$


$\tt{\implies y}$

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$\tt{ {\dfrac{cos^2 \theta - 3cos \theta + 2}{sin^2 \theta}} = 1}$


$\tt{\implies cos^2 \theta - 3cos \theta + 2 = sin^2 \theta}$


${\boxed{\tt{sin^2 \theta = 1 - cos^2 \theta}}}$


$\tt{\implies cos^2 \theta - 3cos \theta + 2 = 1 - cos^2 \theta}$


$\tt{\implies cos^2 \theta - 3cos \theta + 2 - 1 + cos^2 \theta = 0}$


$\tt{\implies 2cos^2 \theta - 3cos^2 \theta + 1 = 0}$


$\tt{\implies 2cos^2 \theta - 2cos \theta - cos \theta + 1 = 0}$


$\tt{\implies 2cos \theta (cos \theta - 1) - 1(cos \theta - 1) = 0}$


$\tt{\implies (2cos \theta - 1)(cos \theta - 1) = 0}$


$\tt{Using \ Zero \ Product \ Rule}$


$\tt{\implies (2cos \theta - 1) = 0 \ and \ (cos \theta - 1) = 0}$


$\tt{\implies cos \theta = {\dfrac{1}{2}} \ and \ cos \theta = 1}$


$\tt{\implies cos \theta = cos60 \degree \ and \ cos \theta = cos0 \degree}$


$\tt{\implies \theta = 60 \degree, \ 0 \degree}$