Saved Bookmarks
| 1. |
1+sin²thita=3sin thita×cos thitaThen prove that tan thita=1or½ |
| Answer» Given, 1 + sin2\xa0θ = 3 sin θ cos θ, then we have to prove that tan θ = 1, or\xa0{tex}\\frac { 1 } { 2 }{/tex}.Now, lets take given trigonometric equation1 + sin2θ = 3 sinθ cosθ\xa0[Dividing by sin2\xa0θ 0n both sides]⇒\xa0{tex}\\frac { 1 } { \\sin ^ { 2 } \\theta } + \\frac { \\sin ^ { 2 } \\theta } { \\sin ^ { 2 } \\theta } = \\frac { 3 \\sin \\theta \\cos \\theta } { \\sin ^ { 2 } \\theta }{/tex}⇒ cosec2θ + 1 = 3 cotθ⇒ 1 + cot2θ + 1 – 3 cotθ = 0⇒ cot2\xa0θ – 3 cot θ + 2 = 0⇒ cot2\xa0θ – 2 cot θ –cot θ + 2 = 0⇒ cot θ(cot θ – 2) –1(cot θ – 2) = 0⇒ (cotθ – 2) (cot θ – 1) = 0⇒ cotθ – 2 = 0 or (cot θ – 1) = 0⇒ cotθ = 2 or cot θ = 1⇒ tanθ =\xa0{tex}\\frac { 1 } { 2 }{/tex}\xa0or tan θ = 1Either tan θ =\xa0{tex}\\frac { 1 } { 2 }{/tex}, or 1which is a required answer. | |