Given, sin(90°−θ)+cosθ=√2cos(90°−θ) ........(1) We know that, sin(90°−θ)cosθ andcos(90°−θ)sinθ Now, Equation (1) can be written as, cosθ+cosθ=√2sinθ (or) 2cosθ=√2sinθ Squaring on bothsides we get, 2cos2θ=sin2θ ⇒ 2(1−sin2θ)⇒sin2θ⇒2−2sin2θ=sinθ ⇒2=3sin2θ⇒2=3(