त्रिकोणमिति

अभ्यास 8.4 Part 6

(vii) `(sinθ-2sin^3θ)/(2cos^3θ-cosθ)=tanθ`

उत्तर:

`LHS: (sinθ-2sin^3θ)/(2cos^3θ-cosθ)`

`=(sinθ(1-2sin^2θ))/(cosθ(2cos^2θ-1))`

`=(sinθ[1-2(1-cos^2&theta)])/(cosθ(2cos^2θ-1))`

`=(sinθ(1-2+2cos^2θ))/(cosθ(2cos^2θ-1))`

`=(sinθ(2cos^2θ-1))/(cosθ(2cos^2θ-1))`

`=(sinθ)/(cosθ)=tanθ=RHS`

(viii) `(sinA+text(cosec)A)^2+(cosA+secA)^2``=7+tan^2A+cot^2A`

उत्तर:

`LHS: (sinA+text(cosec)A)^2+(cosA+secA)^2`

`=sin^2A+text(cosec)^2A+2sinA.text(cosec)A``+cos^2A+sec^2A+2cosA.secA`

`=sin^2A+cos^2A+text(cosec)^2A+sec^2A+2+2`

`=1+2+2+text(cosec)^2A+sec^2A`

`=5+(1)/(sin^2A)+(1)/(cos^2A)`

`=5+(cos^2A+sin^2A)/(sin^2A.cos^2A)`

`=5+(1)/(sin^2A.cos^2A)`

अब;

`RHS: 7+tan^2A+cot^2A`

`=7+(sin^2A)/(cos^2A)+(cos^2A)/(sin^2A)`

`=7+(sin^4A+cos^4A)/(sin^2A.cos^2A)`

`=7+((sin2A+cos^2A)^2-2sin^2A.cos^2A)/(sin^2A.cos^2A)`

`=7+(1+2sin^2A.cos^2A)/(sin^2A.cos^2A)`

`=(7 sin^2A.cos^2A+1-2 sin^2A.cos^2A)/( sin^2A.cos^2A)`

`=(5 sin^2A.cos^2A+1)/( sin^2A.cos^2A)`

`=5+(1)/( sin^2A.cos^2A)=LHS`

इसलिए, LHS = RHS सिद्ध हुआ