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प्रश्न
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
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उत्तर
We have ` ( sec theta + tan theta ) =m ....(i)`
Again ,` ( sec theta - tan theta ) = n .....(ii)`
Now, multiplying (i) and (ii), we get:
`(sec theta + tan theta ) xx ( sec theta - tan theta ) = mn`
` => sec^2 theta - tan^2 theta = mn `
`= > 1= mn [∵ sec^2 theta - tan^2 theta = 1 ]`
∴ mn = 1
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
