If 7 sin2θ+3cos2θ=4, then secθ+cosecθ is equal to
2√3+63
7sin2θ+3cos2θ=4⇒7sin2θ+3(1−sin2θ)=4⇒7sin2θ+3−3sin2θ=4⇒4sin2θ=1⇒sin2θ=14⇒sinθ=12∴θ=30∘∴sec30∘+cosec30∘=(2√3)+2
= 2+2√3√3
=2(1+√3)√3
Multiplying numerator and denominator by √3 , we get
= 2√3+63