Question

log 7 base x + log x base 7 = 5/2​

Answer 1

Given

cos θ + sin θ = √2cos θ

Squaring both side, we get

(cos θ + sin θ)2 = 2cos2θ

cos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θ

sin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θ

sin2θ + 2 × cosθ × sinθ = cos2θ

cos2θ – 2 × cosθ × sinθ = sin2θ

Now adding sin2θ both side, we get

cos2θ -2 × cosθ × sinθ + sin2θ = sin2θ + sin2θ

(cos θ – sin θ)2 = 2sin2θ

cos θ – sin θ = √2sinθ

∴ cos θ – sin θ = √2sinθ