Question
If secθ + tanθ = 3+√10, then the value
of sinθ+cosθ isSolution
secθ + tanθ = 3+√10 ............... (i) (sec²θ - tan²θ) = 1 (secθ + tanθ) (secθ - tanθ) = 1 (3+√10) (secθ - tanθ) = 1 secθ - tanθ = 1/(3+√10) = 1/(√10+3) secθ - tanθ = √10-3 ............ (ii) From (i) and (ii) we get, 2secθ = 2√10 secθ = √10 cosθ = 1/√10 sin²θ + cos²θ = 1 sin²θ + 1/10 = 1 sin²θ = 1- 1/10 = 9/10 sinθ = 3/√10 sinθ + cosθ = 3/√10 + 1/√10 = 4/√10 = 4/√10 × √10/√10 = (2√10)/5
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