If secθ=114, then tanθ2=
13
34
14
54
Step 1 : Apply the formula cos(A)=1−tan2A21+tan2A2
∴secθ=114⇒1cosθ=54⇒cosθ=45⇒45=1−tan2θ21+tan2θ2
Step 2 : Use the cross multiply method
4×1+tan2θ2=5×1-tan2θ24+4tan2θ2=5-5tan2θ24tan2θ2+5tan2θ2=5-49tan2θ2=1tan2θ2=19
Step 3: Take the square root
∴tan2θ2=19=19=13
Hence , the correct option is A.