Webb16 feb. 2024 · Prove that sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1. asked Apr 19, 2024 in Trigonometry by ... 0 votes. 1 answer. Without using trigonometric tables, find the value of (sin^2 10° + sin^2 80°)/(sec^2 20°-cot^2 70°)-3tan80°tan 50° tan 45° tan 10° tan 40°. asked Mar 12, 2024 in Mathematics ... WebbThe sum of the squares of all six roots is thus given by 2(tan2(20 ∘) + tan2(40 ∘) + tan2(80 ∘)). However, if we write these roots as r1,..., r6, then we also have ∑ i r2i = (∑ i ri)2 − 2∑ i < jrirj But ∑iri is the coefficient of x5 in the above equation, namely zero, and ∑i < jrirj is the coefficient of x4, namely − 33.
The value of \( \tan 20^{\circ}+2 \tan 50^{\circ}-\tan 70^{\circ ...
WebbClick here👆to get an answer to your question ️ the value of tan 20 ^∘ + 2 tan 50 ^∘ - tan 70 ^∘ is : Solve Study Textbooks Guides. Join / Login >> Class 11 >> Maths >> Trigonometric Functions >> Trigonometric Functions of Sum and Difference of Two angles >> the value of tan 20 ^∘ + 2 tan 50 ^∘ - t. first bank in immokalee
Prove that tan 50° tan 40° = 2 tan 10° - BYJU
Webb9 feb. 2024 · Best answer. Since, 50° = 10° +40°. ∴ tan 50° = tan (10° + 40°) ∴ tan 10°+tan 40° 1−tan 10° tan 40° t a n 10 ° + t a n 40 ° 1 − t a n 10 ° t a n 40 °. ∴ tan 50° (1 – tan 10° … WebbNow, as 70° + 20° = 90°, ∴ tan70°tan20° = 1, so tan70° = 1 tan20°. tan70° − tan20° − 2tan40° = 1 tan20° − tan20° − 2tan40° = 2(1 − tan220°) 2tan20° − 2tan40° = 2cot40° − … Webbtan 70 = tan ( 20 + 50) ∵ t a n ( A + B) = ( tan A + tan B) 1 – tan A tan B ⇒ tan 70 = ( tan 20 + tan 50) 1 – tan 20 tan 50 ⇒ tan 70 ( 1 – tan 20 tan 50) = tan 20 + tan 50 ⇒ tan 70 – tan 70 tan 20 tan 50 = tan 20 + tan 50 ⇒ tan 70 − tan 50 = tan 20 + tan 50 ( ∵ tan 70 tan 20 = 1) ⇒ tan 70 = tan 20 + 2 tan 50 Hence proved. Suggest Corrections 0 first banking trust