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यदि `tan^(2)theta=1-e^(2)` है तो `sectheta+tan^(3)thetaco s ectheta` का मैं क्या होगा?A. `(2+e^(2))^((3)/(2))`B. `(2-e^(2))^((1)/(2))`C. `(2+e^(2))^((1)/(2))`D. `(2-e^(2))^((3)/(2))`

Answer» Correct Answer - d
`tan^(2)theta=1-e^(2)`
`therefore sec theta+tan^(2)theta.co s ec theta`
`rArrsectheta+tan^(2)theta.tantheta.co s ectheta`
`rArr sec theta+tan^(2)theta.(sintheta)/(cos theta).(1)/(sin theta)`
`rArr sec theta+tan^(@)theta.sec theta`
`rArrsec theta(1+tan^(2)theta)=sqrt(1+tan^(2)theta)`
`(1+tan^(2)theta)`
`rArr(1+tan^(2)theta)^(3//2)=(1+1-e^(2))^(3//2)`
`rArr(2-e^(2))^(3//2)`


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