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Question
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
Solution 1
LHS = ` sin theta (1+ tan theta ) + cos theta ( 1+ cot theta )`
=` sin theta + sin theta xx (sin theta)/(cos theta) + cos theta +cos theta xx (cos theta)/( sin theta)`
= `( cos theta sin ^2 theta + sin^3 theta + cos^2 theta sin theta + cos^3 theta)/(cos theta sin theta)`
=`((sin^3 theta + cos^3 theta)+(cos theta sin ^2 theta + cos ^2 theta sin theta))/(cos theta sin theta)`
=`((sin theta + cos theta )(sin^2 theta - sin theta cos theta + cos ^2 theta )+ sin theta cos theta ( sin theta + cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta )( sin^2 theta + cos^2 theta - sin theta cos theta + sin theta cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta)(1))/(cos theta sin theta)`
= `(sin theta)/(cos theta sin theta) + (cos theta)/( cos theta sin theta)`
=`1/cos theta + 1/ sin theta`
=` sec theta + cosec theta`
=RHS
Solution 2
LHS = ` sin theta (1+ tan theta ) + cos theta ( 1+ cot theta )`
=` sin theta + sin theta xx (sin theta)/(cos theta) + cos theta +cos theta xx (cos theta)/( sin theta)`
= `( cos theta sin ^2 theta + sin^3 theta + cos^2 theta sin theta + cos^3 theta)/(cos theta sin theta)`
=`((sin^3 theta + cos^3 theta)+(cos theta sin ^2 theta + cos ^2 theta sin theta))/(cos theta sin theta)`
=`((sin theta + cos theta )(sin^2 theta - sin theta cos theta + cos ^2 theta )+ sin theta cos theta ( sin theta + cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta )( sin^2 theta + cos^2 theta - sin theta cos theta + sin theta cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta)(1))/(cos theta sin theta)`
= `(sin theta)/(cos theta sin theta) + (cos theta)/( cos theta sin theta)`
=`1/cos theta + 1/ sin theta`
=` sec theta + cosec theta`
=RHS
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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`
