Introduction to Trigonometry — Exercise 8.3

Put c=cotA>0. From cosec²A=1+cot²A, cosecA=√(1+c²). Hence sinA=1/√(1+c²). Also tanA=1/c and cosA=c sinA.

Put s=secA>1. Then cosA=1/s and tanA=√(s²−1), taking the positive root for an acute angle. Obtain the others by quotient or reciprocal.

(i) 9(sec²A−tan²A)=9. (ii) Write s=sinθ,c=cosθ. The product becomes [(c+s+1)/c][(s+c−1)/s]=[(s+c)²−1]/(sc)=2. (iii) (secA+tanA)(1−sinA)=[(1+sinA)/cosA](1−sinA)=cosA. (iv) (1+tan²A)/(1+cot²A)=sec²A/cosec²A=tan²A.

All identities are considered only where the original expressions are defined; the angles are acute. Start with the indicated side and simplify using sin²+cos²=1.

(i) Squared cosecant-minus-cotangent

\[(\csc\theta-\cot\theta)^2=\frac{(1-\cos\theta)^2}{\sin^2\theta}=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}=\frac{1-\cos\theta}{1+\cos\theta}\]

(ii) Sum of the two cosine/sine quotients

\[\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=\frac{\cos^2 A+(1+\sin A)^2}{\cos A(1+\sin A)}=\frac{2(1+\sin A)}{\cos A(1+\sin A)}=2\sec A\]

(iii) Tangent/cotangent fractions

\[\frac{s^2}{c(s-c)}+\frac{c^2}{s(c-s)}=\frac{s^3-c^3}{sc(s-c)}=\frac{s^2+sc+c^2}{sc}=1+\sec\theta\csc\theta\]

Here s=sinθ and c=cosθ. The factor s−c is nonzero wherever the original denominators are nonzero.

(iv) Secant quotient

\[\frac{1+\sec A}{\sec A}=1+\cos A,\qquad\frac{\sin^2 A}{1-\cos A}=\frac{(1-\cos A)(1+\cos A)}{1-\cos A}=1+\cos A\]

(v) Cosine/sine quotient

\[\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\frac{\cot A-1+\csc A}{\cot A+1-\csc A}=\frac{u-1}{1-1/u}=u\]

Set u=cosecA+cotA. Because (cosecA+cotA)(cosecA−cotA)=cosec²A−cot²A=1, the denominator is 1−1/u. Thus u=cosecA+cotA is the required right side.

(vi) Positive square-root identity

\[\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}=\frac{1+\sin A}{\cos A}=\sec A+\tan A\]

The positive root is valid because cosA>0 and 1+sinA>0 for acute A.

(vii) Cubic sine/cosine quotient

\[\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}=\frac{\sin\theta}{\cos\theta}=\tan\theta\]

The bracketed factors agree since 2cos²θ−1=1−2sin²θ. Cancellation applies only when the original denominator is nonzero.

(viii) Sum of two squares

\[(\sin A+\csc A)^2+(\cos A+\sec A)^2=1+4+\csc^2 A+\sec^2 A=7+\cot^2 A+\tan^2 A\]

(ix) Product and reciprocal sum

\[(\csc A-\sin A)(\sec A-\cos A)=\frac{\cos^2 A}{\sin A}\frac{\sin^2 A}{\cos A}=\sin A\cos A\]

Also tanA+cotA=(sin²A+cos²A)/(sinA cosA)=1/(sinA cosA), so its reciprocal is the same result.

(x) The three equal tangent expressions

\[\frac{1+\tan^2 A}{1+\cot^2 A}=\tan^2 A,\qquad\left(\frac{1-t}{1-1/t}\right)^2=(-t)^2=t^2\]

Here t=tanA. The original middle expression excludes t=1; where defined, it equals tan²A.