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Prove that:
(i) cos θ + sin (27θ° + θ) – sin (27θ° – θ)+ cos (18θ° + θ) = θ

(ii) sec(3π/2 - θ)sec(θ - 5π/2) + tan(5π/2 + θ)tan(θ - 3π/2) = - 1

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(i) sin (27θ° + θ) → IV quardant, -ive = -cosθ … (i)

sin (27θ° – θ) → III quardant, -ive = – cos θ …. (ii)

cos (18θ° + θ) → III quardrant, -ive = – cos θ … (iii)

L.H.S. = cos θ + sin (27θ° + θ) – sin (27θ° – θ) + cos (18θ° + 9)

= cos θ – cos θ + cos θ – cos θ [from (i), (ii) and (iii)]

= θ
= R.H.S.

Hence Proved.

(ii) LHS = sec(3π/2 - θ)sec(θ - 5π/2) + tan(5π/2 + θ)tan(θ - 3π/2)

= sec (27θ° – θ) sec (θ – 45θ°)+ tan (45θ° + θ) tan (θ – 27θ°)

= – cosec θ sec (θ – 45θ°)+ tan (36θ° + 9θ° + θ) tan (θ – 27θ°)

= – cosec θ sec (45θ° – θ) – tan (9θ° + θ) tan (27θ° – θ)

= – cosec θ sec (36θ° + 9θ° – θ) – (- cot θ) cot θ

= – cosec θ sec (9θ° – θ) + cot2 θ

= – cosec θ cosec θ + cot2 θ

= – cosec2 θ + cot2 θ

= – 1 (∵ 1 + cot2 θ = cosec2 θ cot2 θ – cosec2 θ = – 1)

R.H.S.

Hence Proved.

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