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solve cot theta - tan theta - costheta +sintheta = 0

solve cot theta - tan theta - costheta +sintheta = 0

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:Hello student, please find answer to your question
cot\theta -tan\theta -cos\theta +sin\theta = 0
cot\theta -tan\theta = cos\theta - sin\theta
\frac{cos\theta }{sin\theta } -\frac{sin\theta }{cos\theta } = cos\theta - sin\theta
\frac{cos^{2}\theta-sin^{2}\theta }{sin\theta.cos\theta } = cos\theta - sin\theta
(cos\theta +sin\theta )(cos\theta-sin\theta) = (cos\theta - sin\theta)sin\theta .cos\theta
(cos\theta -sin\theta )(cos\theta+sin\theta-sin\theta .cos\theta ) = 0
One solution is:
cos\theta -sin\theta= 0
tan\theta = 1
\theta = n\pi +\frac{\pi }{4}
Second solution is
cos\theta+sin\theta-sin\theta .cos\theta = 0
cos\theta+sin\theta=sin\theta .cos\theta
(cos\theta+sin\theta)^{2}=(sin\theta .cos\theta)^{2}
cos^{2}\theta+sin^{2}\theta + 2sin\theta .cos\theta =sin^{2}\theta .cos^{2}\theta
1 + sin2\theta =sin^{2}\theta .cos^{2}\theta
4 + 4sin2\theta =4sin^{2}\theta .cos^{2}\theta
4 + 4sin2\theta =sin^{2}2\theta
sin^{2}2\theta - 4sin2\theta -4 = 0
sin2\theta = \frac{4\pm \sqrt{32}}{2}
sin2\theta = \frac{4\pm 4\sqrt{2}}{2}
sin2\theta = 2\pm 2\sqrt{2}
sin2\theta = 2- 2\sqrt{2}
2\theta = 2n\pi +sin^{-1}(2-2\sqrt{2})
\theta = n\pi +\frac{1}{2}sin^{-1}(2-2\sqrt{2})

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