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| 1. |
Tanθ + tan(90-θ)=secθ + sec(90-θ) |
| Answer» We have to prove that: tan θ + tan (90° – θ) = sec θ sec (90° – θ)Here, LHS = tan θ + tan (90° – θ)= tan θ + cot θ{tex}= \\frac { \\sin \\theta } { \\cos \\theta } + \\frac { \\cos \\theta } { \\sin \\theta }{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta }{/tex}{tex}= \\frac { 1 } { \\sin \\theta \\cos \\theta }{/tex}\xa0[∵ sin2\xa0θ + cos2\xa0θ = 1]= cosec θ·sec θ⇒ LHS = cosec θ·secθRHS = sec θ sec θ (90° - θ)= sec θ cosec θ⇒ LHS = RHSHence, verified. | |