Prove that (cosecθ-sinθ)(secθ-cosθ)=1tanθ+cotθ
Determine the proving of the expression (cosecθ-sinθ)(secθ-cosθ)=1tanθ+cotθ
Solve the L.H.S part:
(cosecθ-sinθ)(secθ-cosθ)=(1sinθ-sinθ)(1cosθ-cosθ)[∵cosecθ=1sinθandsecθ=1cosθ]⇒=(1-sin2θsinθ)(1-cos2θcosθ)[∵sin2θ+cos2θ=1]⇒=cos2θsinθ×sin2θcosθ⇒=cosθsinθ
Solve the R.H.S part:
1tanθ+cotθ=1sinθcosθ+cosθsinθ[∵tanθ=sinθcosθandcotθ=cosθsinθ]⇒=1sin2θ+cos2θcosθsinθ⇒=cosθsinθsin2θ+cos2θ[∵sin2θ+cos2θ=1]⇒=cosθsinθ1⇒=cosθsinθ
Hence,it is proved that (cosecθ-sinθ)(secθ-cosθ)=1tanθ+cotθ
Prove the following trigonometric identities.(i) 1+cosθ+sinθ1+cosθ−sinθ=1+sinθcosθ
(ii) sinθ−cosθ+1sinθ+cosθ−1=1secθ−tanθ
(iii) cosθ−sinθ+1cosθ+sinθ−1=cosecθ+cotθ
(iv) (sinθ+cosθ)(tanθ+cotθ)=secθ+cosecθ
prove that
( sin theta +cos theta ) (cot theta + tan theta) =sec theta +cosec theta.