Taking L.H.S = tan 20° tan 35° tan 45° tan 55° tan 70°
= tan (90° − 70°) tan (90° − 55°) tan 45°tan 55° tan70°
= cot 70°cot 55° tan 45° tan 55° tan 70° [∵ tan (90 – θ) = cot θ]
= (tan 70°cot 70°)(tan 55°cot 55°) tan 45° [∵ tan θ x cot θ = 1]
= 1 × 1 × 1 = 1
Hence proved
Answered by Sakshi | 9 months agoProve the sinθ sin (90° – θ) – cos θ cos (90° – θ) = 0
Prove that sin 48° sec 48° + cos 48° cosec 42° = 2
Evalute the following:
(i) \( \dfrac{sin 20°}{ cos 70°}\)
(ii) \(\dfrac{ cos 19°}{ sin 71°}\)
(iii) \( \dfrac{sin 21°}{ cos 69°}\)
(iv) \( \dfrac{tan 10°}{ cot 80°}\)
(v) \( \dfrac{sec 11°}{ cosec 79°}\)
Evaluate the sec 50° sin 40° + cos 40° cosec 50°