Find the easiest way to solve the question of introduction trigonometry of ncert class 10. How i solve this exercise 8.4 question no. 3(1) easily and quickly.Give me the best solution of this question. Evaluate: (i) (sin263° + sin227°)/(cos217° + cos273°) .
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(sin263° + sin227°)/(cos217° + cos273°)
To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,
= [sin2(90°-27°) + sin227°] / [cos2(90°-73°) + cos273°)]
= (cos227° + sin227°)/(sin227° + cos273°)
= 1/1 =1 (since sin2A + cos2A = 1)
Therefore, (sin263° + sin227°)/(cos217° + cos273°) = 1