Join Examsbook
if $${secθ={a\over b},b≠0,then \ {{1-tan^2θ}\over 2-sin^2θ} }=?$$
5Q:
if $${secθ={a\over b},b≠0,then \ {{1-tan^2θ}\over 2-sin^2θ} }=?$$
- 1$${b^2(2b^2-a^2)}\over {a^2(a^2+b^2)}$$false
- 2$${a^2(2b^2-a^2)}\over {b^2(a^2+b^2)}$$true
- 3$${a^2(2b^2+a^2)}\over {b^2(a^2+b^2)}$$false
- 4$${a^2(2b^2+a^2)}\over {b^2(a^2-b^2)}$$false
- Show AnswerHide Answer
- Workspace