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