(i) tan(-225°) = -(tan 225°)
= -(tan(180° + 45°))
= – tan 45° = – 1
cot(-405°) = -(cot 405°)
= – cot(360° + 45°) [∵ For 360° + 45° no change in T-ratio.]
= -cot 45°
= -1
tan(-765°) = -tan 765°
= -tan(2 x 360° + 45°)
= -tan 45°
= -1
cot 675° = cot (360°+ 315°)
= cot 315°
= cot(360° – 45°)
= -cot 45°
= -1
LHS = tan(-225°) cot(-405°) – tan(-765°) cot(675°)
= (-1) (-1) – (-1) (-1)
= 1 – 1
= 0 = RHS.
Hence proved.
(ii) 2 sin2 \(\frac{\pi}{6}\) + cosec2 \(\frac{7\pi}{6}\) cos2 \(\frac{\pi}{3}\)= \(\frac{3}{2}\)
LHS = 2 sin2 \(\frac{\pi}{6}\) + cosec2 \(\frac{7\pi}{6}\) cos2 \(\frac{\pi}{3}\)= \(\frac{3}{2}\)
[∵ \(\frac{7\pi}{6}\)= 210°, 210° = 180° + 30°. For 180° + 30° no change in T-ratio. 210° lies in 3rd quadrant, cosec θ is negative.]
= 2(sin\(\frac{\pi}{6}\))2 + (cosec (180° + 30°))2 (cos\(\frac{\pi}{3}\))2
= 2(\(\frac{1}{2}\))2 + (-cosec 30°)2.(\(\frac{1}{2}\))2
= 2 x \(\frac{1}{4}+(-2)^2\frac{1}{4}\)
= \(\frac{2}{4}+\frac{4}{4}=\frac{6}{4}\)
= \(\frac{6}{4}\)
= \(\frac{3}{2}\)
= RHS
(iii) sec(\(\frac{3\pi}{2}-\theta\)) = sec (270° – θ) = -cosec θ
[∵ For 270° – θ change T-ratio. So add ‘co’ infront ‘sec’, it becomes ‘cosec’]
sec(θ – \(\frac{5\pi}{2}\)) = sec(-(\(\frac{5\pi}{2}\) - θ))
= sec(\(\frac{5\pi}{2}\) – θ) [∵ sec(-θ) = θ]
= sec(450° – θ)
= sec (360° + (90° – θ))
= sec (90° – θ)
= cosec θ
[∵ For 90° – θ change in T-ratio. So add ‘co’ in front of ‘sec’ it becomes ‘cosec’]
tan(\(\frac{5\pi}{2}\) + θ) = tan(450° + θ)
[∵ For 90° + θ, change in T-ratio. So add ‘co’ in front of ‘tan’ it becomes ‘cot’]
= tan (360° + (90° + θ))
= tan (90° + θ)
= -cot θ
tan(\(\theta-\frac{5\pi}{2}\)) = tan(-(\(\frac{5\pi}{2}\) - θ))
= - tan (\(\frac{5\pi}{2}\) - θ) [∵ tan(-θ) = -tan θ]
= -tan(450° – θ)
= -tan(360° + (90° – θ))
= -tan(90° – θ)
= -cot θ
LHS = sec(\(\frac{3\pi}{2}-\theta\)) sec(\(\theta-\frac{5\pi}{2}\)) + tan(\(\frac{5\pi}{2}+\theta\)) tan(\(\theta-\frac{5\pi}{2}\))
= -cosec θ (cosec θ) + (-cot θ) (-cot θ)
= -cosec2 θ + cot2 θ
= -(1 + cot2 θ) + cot2 θ [∵ 1 + cot2 θ = cosec2 θ]
= -1
= RHS
