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        A quadratic equation whose roots are cosec square thita and sec square thita can be
11 months ago

Arun
22364 Points
							Dear student Equation can be as follows- x² - ( cosec²theta + sec² theta) + cosec ²theta * sec²theta = 0 x² - (tan theta + cot theta)² x + 1/sin²theta * cos²theta = 0 RegardsArun (askIITians forum expert)

11 months ago
kkbisht
90 Points
							As you know if  $\alpha$,$\beta$ are the roots of a quadratic equation then the equation formed is :   x2 -( $\alpha$ +$\beta$)x + $\alpha$.$\beta$=0 => x2 -( cosec2 $\theta$ + sec2$\theta$)x +cosec2 $\theta$ . sec2$\theta$=0 => x2 -( 1/sin2$\theta$ + 1/cos2 $\theta$)x + 1/sin2$\theta$ .1/cos2 $\theta$=0=>x2 -( sin2$\theta$ + cos2 $\theta$). (1/sin2$\theta$ .cos2 $\theta$)x + 1/sin2$\theta$ .1/cos2 $\theta$=0 x2 -( 1/(sin2$\theta$ .cos2 $\theta$)x + 1/sin2$\theta$ .1/cos2 $\theta$=0=> (sin2$\theta$ .cos2 $\theta$) x2  -x  +1=0kkbisht

11 months ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions