Numerical: trigonometric identities MCQ — 3 Practice Questions with Answers

Explanation

Given sin 23° = a/b, cos 23° = √(b^2-a^2)/b. Also, sin 67° = cos 23° because the angles are complementary. Therefore sec 23° - sin 67° = 1/cos 23° - cos 23° = b/√(b^2-a^2) - √(b^2-a^2)/b. Simplifying gives a^2/(b√(b^2-a^2)). The expressions with a in the numerator, with b^2+a^2, or without the square-root denominator do not follow from this identity.

Explanation

From \(\cosec \theta = x/y\), \(\sin \theta = y/x\). Therefore \(\cos \theta = \sqrt{x^2-y^2}/x\), so \(\cot \theta = \sqrt{x^2-y^2}/y\) and \(\tan \theta = y/\sqrt{x^2-y^2}\). Put \(u = \sqrt{x^2-y^2}/y\). Then \(\tan \theta = 1/u\), and \((\sqrt{3}u+1)/(1/u+\sqrt{3}) = u\). Thus the value is \(\sqrt{x^2-y^2}/y\). The forms with denominator \(x\) or with \(x^2+y^2\) misuse the Pythagorean relation.

Explanation

Given cosecθ + cotθ = m. Using the identity (cosecθ + cotθ)(cosecθ - cotθ) = cosec^2θ - cot^2θ = 1, we get cosecθ - cotθ = 1/m. Adding the two equations gives 2cosecθ = m + 1/m, while subtracting them gives 2cotθ = m - 1/m. Dividing cotθ by cosecθ gives cosθ = (m - 1/m)/(m + 1/m) = (m^2 - 1)/(m^2 + 1). Thus sinθ, cosecθ and tanθ do not match the derived expression.