Class 9: Complementary Angles

$\displaystyle \textbf{Exercise - 25(A)}$

$\displaystyle \textbf{Question 1: }\text{Evaluate:}$
$\displaystyle \text{(i) }\frac{\cos22^\circ}{\sin68^\circ}\qquad \text{(ii) }\frac{\tan47^\circ}{\cot43^\circ}\qquad \text{(iii) }\frac{\sec75^\circ}{\mathrm{cosec}\,15^\circ}$
$\displaystyle \text{(iv) }\frac{\cos55^\circ}{\sin35^\circ}+\frac{\cot35^\circ}{\tan55^\circ}\qquad \text{(v) }\sin^2 40^\circ-\cos^2 50^\circ$
$\displaystyle \text{(vi) }\sec^2 18^\circ-\mathrm{cosec}^2 72^\circ$
$\displaystyle \text{(vii) }\sin15^\circ\cos15^\circ-\cos75^\circ\sin75^\circ$
$\displaystyle \text{(viii) }\sin42^\circ\sin48^\circ-\cos42^\circ\cos48^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\frac{\cos22^\circ}{\sin68^\circ}$
$\displaystyle =\frac{\cos22^\circ}{\cos(90^\circ-68^\circ)}$
$\displaystyle =\frac{\cos22^\circ}{\cos22^\circ}=1$

$\displaystyle \text{(ii) }\frac{\tan47^\circ}{\cot43^\circ}$
$\displaystyle =\frac{\tan47^\circ}{\tan(90^\circ-43^\circ)}$
$\displaystyle =\frac{\tan47^\circ}{\tan47^\circ}=1$

$\displaystyle \text{(iii) }\frac{\sec75^\circ}{\mathrm{cosec}\,15^\circ}$
$\displaystyle =\frac{\sec75^\circ}{\sec(90^\circ-15^\circ)}$
$\displaystyle =\frac{\sec75^\circ}{\sec75^\circ}=1$

$\displaystyle \text{(iv) }\frac{\cos55^\circ}{\sin35^\circ}+\frac{\cot35^\circ}{\tan55^\circ}$
$\displaystyle =\frac{\cos55^\circ}{\cos(90^\circ-35^\circ)} +\frac{\cot35^\circ}{\cot(90^\circ-55^\circ)}$
$\displaystyle =\frac{\cos55^\circ}{\cos55^\circ} +\frac{\cot35^\circ}{\cot35^\circ}$
$\displaystyle =1+1=2$

$\displaystyle \text{(v) }\sin^2 40^\circ-\cos^2 50^\circ$
$\displaystyle =\sin^2 40^\circ-\sin^2(90^\circ-50^\circ)$
$\displaystyle =\sin^2 40^\circ-\sin^2 40^\circ=0$

$\displaystyle \text{(vi) }\sec^2 18^\circ-\mathrm{cosec}^2 72^\circ$
$\displaystyle =\sec^2 18^\circ-\sec^2(90^\circ-72^\circ)$
$\displaystyle =\sec^2 18^\circ-\sec^2 18^\circ=0$

$\displaystyle \text{(vii) }\sin15^\circ\cos15^\circ-\cos75^\circ\sin75^\circ$
$\displaystyle =\sin15^\circ\cos15^\circ -\sin15^\circ\cos15^\circ$
$\displaystyle =0$

$\displaystyle \text{(viii) }\sin42^\circ\sin48^\circ-\cos42^\circ\cos48^\circ$
$\displaystyle =\sin42^\circ\cos42^\circ -\cos42^\circ\sin42^\circ$
$\displaystyle =0$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Evaluate:}$
$\displaystyle \text{(i) }\sin(90^\circ-A)\sin A-\cos(90^\circ-A)\cos A$
$\displaystyle \text{(ii) }\sin^2 35^\circ-\cos^2 55^\circ$
$\displaystyle \text{(iii) }\frac{\cot54^\circ}{\tan36^\circ}+\frac{\tan20^\circ}{\cot70^\circ}-2$
$\displaystyle \text{(iv) }\frac{2\tan53^\circ}{\cot37^\circ}-\frac{\cot80^\circ}{\tan10^\circ}$
$\displaystyle \text{(v) }\cos^2 25^\circ-\sin^2 65^\circ-\tan^2 45^\circ$
$\displaystyle \text{(vi) }\left(\frac{\sin77^\circ}{\cos13^\circ}\right)^2+ \left(\frac{\cos77^\circ}{\sin13^\circ}\right)^2-2\cos^2 45^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\sin(90^\circ-A)\sin A-\cos(90^\circ-A)\cos A$
$\displaystyle =\cos A\sin A-\sin A\cos A$
$\displaystyle =0$

$\displaystyle \text{(ii) }\sin^2 35^\circ-\cos^2 55^\circ$
$\displaystyle =\sin^2 35^\circ-\sin^2(90^\circ-55^\circ)$
$\displaystyle =\sin^2 35^\circ-\sin^2 35^\circ=0$

$\displaystyle \text{(iii) }\frac{\cot54^\circ}{\tan36^\circ}+ \frac{\tan20^\circ}{\cot70^\circ}-2$
$\displaystyle =\frac{\tan(90^\circ-54^\circ)}{\tan36^\circ}+ \frac{\tan20^\circ}{\tan(90^\circ-70^\circ)}-2$
$\displaystyle =\frac{\tan36^\circ}{\tan36^\circ}+ \frac{\tan20^\circ}{\tan20^\circ}-2$
$\displaystyle =1+1-2=0$

$\displaystyle \text{(iv) }\frac{2\tan53^\circ}{\cot37^\circ}- \frac{\cot80^\circ}{\tan10^\circ}$
$\displaystyle =\frac{2\tan53^\circ}{\tan(90^\circ-37^\circ)}- \frac{\tan(90^\circ-80^\circ)}{\tan10^\circ}$
$\displaystyle =\frac{2\tan53^\circ}{\tan53^\circ}- \frac{\tan10^\circ}{\tan10^\circ}$
$\displaystyle =2-1=1$

$\displaystyle \text{(v) }\cos^2 25^\circ-\sin^2 65^\circ-\tan^2 45^\circ$
$\displaystyle =\cos^2 25^\circ-\cos^2(90^\circ-65^\circ)-1$
$\displaystyle =\cos^2 25^\circ-\cos^2 25^\circ-1$
$\displaystyle =-1$

$\displaystyle \text{(vi) }\left(\frac{\sin77^\circ}{\cos13^\circ}\right)^2+ \left(\frac{\cos77^\circ}{\sin13^\circ}\right)^2-2\cos^2 45^\circ$
$\displaystyle =\left(\frac{\sin77^\circ}{\sin(90^\circ-13^\circ)}\right)^2+ \left(\frac{\cos77^\circ}{\cos(90^\circ-13^\circ)}\right)^2-2\left(\frac{1}{\sqrt2}\right)^2$
$\displaystyle =\left(\frac{\sin77^\circ}{\sin77^\circ}\right)^2+ \left(\frac{\cos77^\circ}{\cos77^\circ}\right)^2-2\times\frac{1}{2}$
$\displaystyle =1+1-1=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Show that:}$
$\displaystyle \text{(i) }\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ=1$
$\displaystyle \text{(ii) }\sin42^\circ\sec48^\circ+\cos42^\circ\mathrm{cosec}\,48^\circ=2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) L.H.S.}=\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$
$\displaystyle =\tan10^\circ\tan15^\circ\cot15^\circ\cot10^\circ$
$\displaystyle =\tan10^\circ\cot10^\circ\times\tan15^\circ\cot15^\circ$
$\displaystyle =1\times1=1=\text{R.H.S.}$
$\displaystyle \therefore \text{Hence proved.}$

$\displaystyle \text{(ii) L.H.S.}=\sin42^\circ\sec48^\circ+ \cos42^\circ\mathrm{cosec}\,48^\circ$
$\displaystyle =\sin42^\circ\mathrm{cosec}\,42^\circ+ \cos42^\circ\sec42^\circ$
$\displaystyle =1+1=2=\text{R.H.S.}$
$\displaystyle \therefore \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Express each of the following in terms of angles between }$
$\displaystyle 0^\circ\text{ and }45^\circ\text{:}$
$\displaystyle \text{(i) }\sin59^\circ+\tan63^\circ$
$\displaystyle \text{(ii) }\mathrm{cosec}\,68^\circ+\cot72^\circ$
$\displaystyle \text{(iii) }\cos74^\circ+\sec67^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\sin59^\circ+\tan63^\circ$
$\displaystyle =\cos(90^\circ-59^\circ)+\cot(90^\circ-63^\circ)$
$\displaystyle =\cos31^\circ+\cot27^\circ$

$\displaystyle \text{(ii) }\mathrm{cosec}\,68^\circ+\cot72^\circ$
$\displaystyle =\sec(90^\circ-68^\circ)+\tan(90^\circ-72^\circ)$
$\displaystyle =\sec22^\circ+\tan18^\circ$

$\displaystyle \text{(iii) }\cos74^\circ+\sec67^\circ$
$\displaystyle =\sin(90^\circ-74^\circ)+\mathrm{cosec}(90^\circ-67^\circ)$
$\displaystyle =\sin16^\circ+\mathrm{cosec}\,23^\circ$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{For triangle }ABC,\text{ show that:}$
$\displaystyle \text{(i) }\sin\frac{A+B}{2}=\cos\frac{C}{2}$
$\displaystyle \text{(ii) }\tan\frac{B+C}{2}=\cot\frac{A}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ABC,\quad A+B+C=180^\circ$

$\displaystyle \text{(i) }A+B=180^\circ-C$
$\displaystyle \Rightarrow \frac{A+B}{2}=90^\circ-\frac{C}{2}$
$\displaystyle \therefore \sin\frac{A+B}{2}=\sin\left(90^\circ-\frac{C}{2}\right)$
$\displaystyle =\cos\frac{C}{2}$
$\displaystyle \therefore \text{Hence proved.}$

$\displaystyle \text{(ii) }B+C=180^\circ-A$
$\displaystyle \Rightarrow \frac{B+C}{2}=90^\circ-\frac{A}{2}$
$\displaystyle \therefore \tan\frac{B+C}{2}=\tan\left(90^\circ-\frac{A}{2}\right)$
$\displaystyle =\cot\frac{A}{2}$
$\displaystyle \therefore \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Evaluate:}$
$\displaystyle \text{(i) }3\frac{\sin72^\circ}{\cos18^\circ}-\frac{\sec32^\circ}{\mathrm{cosec}\,58^\circ}$
$\displaystyle \text{(ii) }3\cos80^\circ\,\mathrm{cosec}\,10^\circ+2\sin59^\circ\sec31^\circ$
$\displaystyle \text{(iii) }\frac{\sin80^\circ}{\cos10^\circ}+\sin59^\circ\sec31^\circ$
$\displaystyle \text{(iv) }\tan(55^\circ-A)-\cot(35^\circ+A)$
$\displaystyle \text{(v) }\mathrm{cosec}(65^\circ+A)-\sec(25^\circ-A)$
$\displaystyle \text{(vi) }2\frac{\tan57^\circ}{\cot33^\circ}-\frac{\cot70^\circ}{\tan20^\circ}-\sqrt2\cos45^\circ$
$\displaystyle \text{(vii) }\frac{\cot^2 41^\circ}{\tan^2 49^\circ}-2\frac{\sin^2 75^\circ}{\cos^2 15^\circ}$
$\displaystyle \text{(viii) }\frac{\cos70^\circ}{\sin20^\circ}+\frac{\cos59^\circ}{\sin31^\circ}-8\sin^2 30^\circ$
$\displaystyle \text{(ix) }14\sin30^\circ+6\cos60^\circ-5\tan45^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }3\frac{\sin72^\circ}{\cos18^\circ}-\frac{\sec32^\circ}{\mathrm{cosec}\,58^\circ}$
$\displaystyle =3\frac{\cos18^\circ}{\cos18^\circ}-\frac{\sec32^\circ}{\sec32^\circ}$
$\displaystyle =3-1=2$

$\displaystyle \text{(ii) }3\cos80^\circ\,\mathrm{cosec}\,10^\circ+2\sin59^\circ\sec31^\circ$
$\displaystyle =3\sin10^\circ\,\mathrm{cosec}\,10^\circ+2\cos31^\circ\sec31^\circ$
$\displaystyle =3\times1+2\times1=5$

$\displaystyle \text{(iii) }\frac{\sin80^\circ}{\cos10^\circ}+\sin59^\circ\sec31^\circ$
$\displaystyle =\frac{\cos10^\circ}{\cos10^\circ}+\cos31^\circ\sec31^\circ$
$\displaystyle =1+1=2$

$\displaystyle \text{(iv) }\tan(55^\circ-A)-\cot(35^\circ+A)$
$\displaystyle =\tan(55^\circ-A)-\tan\{90^\circ-(35^\circ+A)\}$
$\displaystyle =\tan(55^\circ-A)-\tan(55^\circ-A)=0$

$\displaystyle \text{(v) }\mathrm{cosec}(65^\circ+A)-\sec(25^\circ-A)$
$\displaystyle =\sec\{90^\circ-(65^\circ+A)\}-\sec(25^\circ-A)$
$\displaystyle =\sec(25^\circ-A)-\sec(25^\circ-A)=0$

$\displaystyle \text{(vi) }2\frac{\tan57^\circ}{\cot33^\circ}-\frac{\cot70^\circ}{\tan20^\circ}-\sqrt2\cos45^\circ$
$\displaystyle =2\frac{\tan57^\circ}{\tan57^\circ}-\frac{\tan20^\circ}{\tan20^\circ}-\sqrt2\left(\frac{1}{\sqrt2}\right)$
$\displaystyle =2-1-1=0$

$\displaystyle \text{(vii) }\frac{\cot^2 41^\circ}{\tan^2 49^\circ}-2\frac{\sin^2 75^\circ}{\cos^2 15^\circ}$
$\displaystyle =\frac{\cot^2 41^\circ}{\cot^2 41^\circ}-2\frac{\cos^2 15^\circ}{\cos^2 15^\circ}$
$\displaystyle =1-2=-1$

$\displaystyle \text{(viii) }\frac{\cos70^\circ}{\sin20^\circ}+\frac{\cos59^\circ}{\sin31^\circ}-8\sin^2 30^\circ$
$\displaystyle =\frac{\sin20^\circ}{\sin20^\circ}+\frac{\sin31^\circ}{\sin31^\circ}-8\left(\frac12\right)^2$
$\displaystyle =1+1-8\times\frac14$
$\displaystyle =2-2=0$

$\displaystyle \text{(ix) }14\sin30^\circ+6\cos60^\circ-5\tan45^\circ$
$\displaystyle =14\left(\frac12\right)+6\left(\frac12\right)-5(1)$
$\displaystyle =7+3-5=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{A triangle }ABC\text{ is right angled at }B;\text{ find the value of}$
$\displaystyle \frac{\sec A\cdot\sin C-\tan A\cdot\tan C}{\sin B}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }\triangle ABC\text{ is right angled at }B,$
$\displaystyle A+C=90^\circ\quad\text{and}\quad B=90^\circ$
$\displaystyle \therefore \sin C=\cos A,\quad \tan C=\cot A,\quad \sin B=1$
$\displaystyle \frac{\sec A\cdot\sin C-\tan A\cdot\tan C}{\sin B}$
$\displaystyle =\frac{\sec A\cdot\cos A-\tan A\cdot\cot A}{1}$
$\displaystyle =1-1=0$
$\displaystyle \therefore \text{The value of the given expression is }0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{In each case, given below, find the value of angle }A,\text{ where}$
$\displaystyle 0^\circ\leq A\leq90^\circ.$
$\displaystyle \text{(i) }\sin(90^\circ-3A)\cdot\mathrm{cosec}\,42^\circ=1$
$\displaystyle \text{(ii) }\cos(90^\circ-A)\cdot\sec77^\circ=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\sin(90^\circ-3A)\cdot\mathrm{cosec}\,42^\circ=1$
$\displaystyle \cos3A\cdot\frac{1}{\sin42^\circ}=1$
$\displaystyle \Rightarrow \cos3A=\sin42^\circ$
$\displaystyle \Rightarrow \cos3A=\cos(90^\circ-42^\circ)$
$\displaystyle \Rightarrow \cos3A=\cos48^\circ$
$\displaystyle \therefore 3A=48^\circ$
$\displaystyle \therefore A=16^\circ$

$\displaystyle \text{(ii) }\cos(90^\circ-A)\cdot\sec77^\circ=1$
$\displaystyle \sin A\cdot\frac{1}{\cos77^\circ}=1$
$\displaystyle \Rightarrow \sin A=\cos77^\circ$
$\displaystyle \Rightarrow \sin A=\sin(90^\circ-77^\circ)$
$\displaystyle \Rightarrow \sin A=\sin13^\circ$
$\displaystyle \therefore A=13^\circ$
$\displaystyle \\$