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If Sec θ + Tan θ = M, Show that M 2 − 1 M 2 + 1 = Sin θ - Mathematics

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प्रश्न

If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`

योग

उत्तर

`(m^2 - 1)/(m^2 + 1)`

⇒ `((sectheta + tantheta)^2 - (sec^2theta - tan^2theta))/((sectheta + tantheta)^2 + (sec^2theta - tan^2theta))`

⇒ `(sec^2theta + tan^2theta + 2sectheta tantheta - sec^2theta + tan^2theta)/(sec^2theta + tan^2theta + 2sectheta tantheta + sec^2theta - tan^2theta)`

⇒ `(2tantheta(tantheta + sectheta))/(2sectheta(tantheta + sectheta))`

⇒ `tantheta/sectheta = sintheta/(costheta sectheta)`   `(∵ tantheta = sintheta/costheta)`

⇒ `sintheta/(costheta xx 1/costheta) = sin theta`  `(∵ costheta = 1/sectheta)`

Hence, proved

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