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Question

Prove that
(i) cos (2π+θ) cosec (2π+θ) tan (π/2+θ)sec(π/2+θ)cosθ cot(π+θ)=1
(ii) cosec(90°+θ)+cot(450°+θ)cosec(90°-θ)+tan(180°-θ)+tan(180°+θ)+sec(180°-θ)tan(360°+θ)-sec(-θ)=2
(iii) sin(180°+θ) cos(90°+θ) tan(270°-θ) cot(360°-θ)sin(360°-θ) cos(360°+θ) cosec(-θ) sin(270°+θ)=1
(iv) 1+cotθ-secπ2+θ1+cotθ+secπ2+θ=2cotθ
(v) tan (90°-θ) sec(180°-θ) sin(-θ)sin(180°+θ) cot(360°-θ) cosec(90°-θ)=1

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Solution

i LHS = cos2π+θ cosec2π+θ tanπ2+θsec π2+θ cos θ cot π+θ =cos θ cosec θ - cot θ-cosec θcos θ cot θ =-cos θ cosec θ cot θ-cosec θ cos θ cot θ = 1 =RHSHence proved.

ii LHS =cosec 90°+θ+cot 450°+θcosec 90°-θ+tan 180°-θ + tan 180°+θ+sec 180°-θtan 360°+θ-sec -θ =cosec90°+θ+cot450°+θcosec 90°-θ+tan180°-θ + tan 180°+θ+sec 180°-θtan 360°+θ-sec -θ =cosec90°+θ+cot 90°×5+θcosec90°-θ+tan 90°×2-θ + tan 90°×2+θ+sec 90°×2-θtan90°×4+θ-sec-θ =sec θ+cot 90°×5+θcosec90°-θ+tan 90°×2-θ + tan 90°×2+θ+sec 90°×2-θtan 90°×4+θ-sec -θ =sec θ-tan θsec θ-tan θ + tan θ-sec θtan θ-sec θ =1 + 1 =2 =RHSHence proved.
iii LHS =sin 180°+θcos90°+θ tan 270°-θcot 360°-θsin 360°-θcos360°+θcosec-θ sin 270°+θ =sin 90×2°+θcos90°×1+θ tan90°×3-θ cot90°×4-θsin90°×4-θcos90°×4+θ cosec -θ sin 90°×3+θ =-sin θ -sin θ cot θ-cot θ-sin θ cos θ -cosec θ-cos θ =sin2θ cot2 θsin θ cosec θ cos θ cos θ =sin2θ ×cos2θsin2θsin θ ×1sin θ×cos2θ =cos2 θcos2 θ =1 =RHSHence proved.

iv LHS = 1+cot θ-secπ2+θ1+cot θ+secπ2+θ = 1+cot θ--cosec θ 1+cot θ+-cosecθ = 1+cot θ+cosec θ 1+cot θ-cosec θ = 1+cot θ+cosec θ 1+cot θ-cosec θ = 1+cotθ+cosec θ 1+cot θ-cosecθ =1+cot θ2 - cosec θ2

=1+cot2θ+2cot θ-cosec2 θ =2 cot θ 1+cot2θ=cosec2θ =RHS Hence proved.



v LHS =tan 90°-θ sec 180°-θ sin -θsin180°+θcot 360°-θcosec 90°-θ =tan 90°×1-θsec 90°×2-θsin -θsin 90°×2+θcot 90°×4-θcosec 90°×1-θ =cot θ-sec θ-sin θ-sin θ-cot θ sec θ =cot θ sec θ sin θsin θ cot θ sec θ =cos θsin θ×1cos θ×sin θsin θ×cos θsin θ×1cos θ =11 =1 =RHSHence proved.

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