Class 9: Trigonometric Ratios

$\displaystyle \textbf{Question 1: }\text{Evaluate each of the following:}$
$\displaystyle \text{(i) }\mathrm{cosec}\,30^\circ+\cot45^\circ$
$\displaystyle \text{(ii) }\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ$
$\displaystyle \text{(iii) }\tan30^\circ\sec45^\circ+\tan60^\circ\sec30^\circ$
$\displaystyle \text{(iv) }\sin30^\circ\sin45^\circ+\cos30^\circ\cos45^\circ$
$\displaystyle \text{(v) }\frac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}$
$\displaystyle \text{(vi) }\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$\displaystyle \text{(vii) }\frac{\sin60^\circ}{\cos^2 45^\circ}-\cot30^\circ+15\cos90^\circ$
$\displaystyle \text{(viii) }\frac{5\sin^2 30^\circ+\cos^2 45^\circ-4\tan^2 30^\circ}{2\sin30^\circ\cos30^\circ+\tan45^\circ}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\mathrm{cosec}\,30^\circ+\cot45^\circ$
$\displaystyle =2+1=3$

$\displaystyle \text{(ii) }\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ$
$\displaystyle =\frac{\sqrt3}{2}\times\frac1{\sqrt2}-\frac12\times\frac1{\sqrt2}$
$\displaystyle =\frac{\sqrt3-1}{2\sqrt2}$

$\displaystyle \text{(iii) }\tan30^\circ\sec45^\circ+\tan60^\circ\sec30^\circ$
$\displaystyle =\frac1{\sqrt3}\times\sqrt2+\sqrt3\times\frac2{\sqrt3}$
$\displaystyle =\frac{\sqrt2}{\sqrt3}+2$
$\displaystyle =\frac{\sqrt2+2\sqrt3}{\sqrt3}$

$\displaystyle \text{(iv) }\sin30^\circ\sin45^\circ+\cos30^\circ\cos45^\circ$
$\displaystyle =\frac12\times\frac1{\sqrt2}+\frac{\sqrt3}{2}\times\frac1{\sqrt2}$
$\displaystyle =\frac{\sqrt3+1}{2\sqrt2}$

$\displaystyle \text{(v) }\frac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}$
$\displaystyle =\frac{\left(\frac1{\sqrt2}\right)^2+\left(\frac1{\sqrt2}\right)^2}{(\sqrt3)^2}$
$\displaystyle =\frac{\frac12+\frac12}{3}=\frac13$

$\displaystyle \text{(vi) }\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$\displaystyle =\frac{\frac12-1+2\times1}{\frac1{\sqrt3}\times\sqrt3}$
$\displaystyle =\frac{\frac32}{1}=\frac32$

$\displaystyle \text{(vii) }\frac{\sin60^\circ}{\cos^2 45^\circ}-\cot30^\circ+15\cos90^\circ$
$\displaystyle =\frac{\frac{\sqrt3}{2}}{\left(\frac1{\sqrt2}\right)^2}-\sqrt3+15\times0$
$\displaystyle =\frac{\frac{\sqrt3}{2}}{\frac12}-\sqrt3$
$\displaystyle =\sqrt3-\sqrt3=0$

$\displaystyle \text{(viii) }\frac{5\sin^2 30^\circ+\cos^2 45^\circ-4\tan^2 30^\circ}{2\sin30^\circ\cos30^\circ+\tan45^\circ}$
$\displaystyle =\frac{5\left(\frac12\right)^2+\left(\frac1{\sqrt2}\right)^2-4\left(\frac1{\sqrt3}\right)^2}{2\times\frac12\times\frac{\sqrt3}{2}+1}$
$\displaystyle =\frac{\frac54+\frac12-\frac43}{\frac{\sqrt3}{2}+1}$
$\displaystyle =\frac{\frac5{12}}{\frac{\sqrt3+2}{2}}$
$\displaystyle =\frac{5}{6(\sqrt3+2)}$
$\displaystyle =\frac56(2-\sqrt3)$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the value of }\theta\text{ in each of the following:}$
$\displaystyle \text{(i) }2\sin\theta=\sqrt3$
$\displaystyle \text{(ii) }2\cos3\theta=1$
$\displaystyle \text{(iii) }\sqrt3\tan2\theta-3=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }2\sin\theta=\sqrt3$
$\displaystyle \Rightarrow \sin\theta=\frac{\sqrt3}{2}$
$\displaystyle \Rightarrow \sin\theta=\sin60^\circ$
$\displaystyle \therefore \theta=60^\circ$

$\displaystyle \text{(ii) }2\cos3\theta=1$
$\displaystyle \Rightarrow \cos3\theta=\frac12$
$\displaystyle \Rightarrow \cos3\theta=\cos60^\circ$
$\displaystyle \Rightarrow 3\theta=60^\circ$
$\displaystyle \therefore \theta=20^\circ$

$\displaystyle \text{(iii) }\sqrt3\tan2\theta-3=0$
$\displaystyle \Rightarrow \sqrt3\tan2\theta=3$
$\displaystyle \Rightarrow \tan2\theta=\frac3{\sqrt3}=\sqrt3$
$\displaystyle \Rightarrow \tan2\theta=\tan60^\circ$
$\displaystyle \Rightarrow 2\theta=60^\circ$
$\displaystyle \therefore \theta=30^\circ$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Evaluate / Prove the following:}$
$\displaystyle \text{(i) }\frac{\cos37^\circ}{\sin53^\circ}$
$\displaystyle \text{(ii) }\frac{\tan54^\circ}{\cot36^\circ}$
$\displaystyle \text{(iii) }\sin39^\circ-\cos51^\circ$
$\displaystyle \text{(iv) }\cot34^\circ-\tan56^\circ$
$\displaystyle \text{(v) }\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\,\mathrm{cosec}\,31^\circ$
$\displaystyle \text{(vi) }\sec50^\circ\sin40^\circ+\cos40^\circ\,\mathrm{cosec}\,50^\circ$
$\displaystyle \text{(vii) }\left(\frac{\sin35^\circ}{\cos55^\circ}\right)^2+\left(\frac{\cos55^\circ}{\sin35^\circ}\right)^2-2\cos60^\circ$
$\displaystyle \text{(viii) }\cos(40^\circ-\theta)-\sin(50^\circ+\theta)$
$\displaystyle \qquad+\frac{\cos^2 40^\circ+\cos^2 50^\circ}{\sin^2 40^\circ+\sin^2 50^\circ}$
$\displaystyle \text{(ix) }\cot12^\circ\cot38^\circ\cot52^\circ\cot60^\circ\cot78^\circ$
$\displaystyle \text{(x) }\tan5^\circ\tan25^\circ\tan30^\circ\tan65^\circ\tan85^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\frac{\cos37^\circ}{\sin53^\circ}$
$\displaystyle =\frac{\cos(90^\circ-53^\circ)}{\sin53^\circ}$
$\displaystyle =\frac{\sin53^\circ}{\sin53^\circ}=1$

$\displaystyle \text{(ii) }\frac{\tan54^\circ}{\cot36^\circ}$
$\displaystyle =\frac{\tan(90^\circ-36^\circ)}{\cot36^\circ}$
$\displaystyle =\frac{\cot36^\circ}{\cot36^\circ}=1$

$\displaystyle \text{(iii) }\sin39^\circ-\cos51^\circ$
$\displaystyle =\sin39^\circ-\cos(90^\circ-39^\circ)$
$\displaystyle =\sin39^\circ-\sin39^\circ=0$

$\displaystyle \text{(iv) }\cot34^\circ-\tan56^\circ$
$\displaystyle =\cot34^\circ-\tan(90^\circ-34^\circ)$
$\displaystyle =\cot34^\circ-\cot34^\circ=0$

$\displaystyle \text{(v) }\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\,\mathrm{cosec}\,31^\circ$
$\displaystyle =\frac{\cos(90^\circ-10^\circ)}{\sin10^\circ}+\cos59^\circ\,\mathrm{cosec}(90^\circ-59^\circ)$
$\displaystyle =\frac{\sin10^\circ}{\sin10^\circ}+\cos59^\circ\sec59^\circ$
$\displaystyle =1+1=2$

$\displaystyle \text{(vi) }\sec50^\circ\sin40^\circ+\cos40^\circ\,\mathrm{cosec}\,50^\circ$
$\displaystyle =\sec50^\circ\sin(90^\circ-50^\circ)+\cos40^\circ\,\mathrm{cosec}(90^\circ-40^\circ)$
$\displaystyle =\sec50^\circ\cos50^\circ+\cos40^\circ\sec40^\circ$
$\displaystyle =1+1=2$

$\displaystyle \text{(vii) }\left(\frac{\sin35^\circ}{\cos55^\circ}\right)^2+\left(\frac{\cos55^\circ}{\sin35^\circ}\right)^2-2\cos60^\circ$
$\displaystyle =\left(\frac{\sin(90^\circ-55^\circ)}{\cos55^\circ}\right)^2+\left(\frac{\cos(90^\circ-35^\circ)}{\sin35^\circ}\right)^2-2\cos60^\circ$
$\displaystyle =\left(\frac{\cos55^\circ}{\cos55^\circ}\right)^2+\left(\frac{\sin35^\circ}{\sin35^\circ}\right)^2-2\times\frac12$
$\displaystyle =1+1-1=1$

$\displaystyle \text{(viii) }\cos(40^\circ-\theta)-\sin(50^\circ+\theta)$
$\displaystyle \qquad+\frac{\cos^2 40^\circ+\cos^2 50^\circ}{\sin^2 40^\circ+\sin^2 50^\circ}$
$\displaystyle =\sin(50^\circ+\theta)-\sin(50^\circ+\theta)$
$\displaystyle \qquad+\frac{\cos^2 40^\circ+\sin^2 40^\circ}{\sin^2 40^\circ+\cos^2 40^\circ}$
$\displaystyle =0+1=1$

$\displaystyle \text{(ix) }\cot12^\circ\cot38^\circ\cot52^\circ\cot60^\circ\cot78^\circ$
$\displaystyle =\tan78^\circ\tan52^\circ\cot52^\circ\cot60^\circ\cot78^\circ$
$\displaystyle =\cot60^\circ$
$\displaystyle =\frac1{\sqrt3}$

$\displaystyle \text{(x) }\tan5^\circ\tan25^\circ\tan30^\circ\tan65^\circ\tan85^\circ$
$\displaystyle =\cot85^\circ\cot65^\circ\tan30^\circ\tan65^\circ\tan85^\circ$
$\displaystyle =\tan30^\circ$
$\displaystyle =\frac1{\sqrt3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Express each of the following in terms of trigonometric ratios of angles between}$
$\displaystyle 0^\circ\text{ and }45^\circ.$
$\displaystyle \text{(i) }\sin85^\circ+\mathrm{cosec}\,85^\circ$
$\displaystyle \text{(ii) }\mathrm{cosec}\,69^\circ+\cot69^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\sin85^\circ+\mathrm{cosec}\,85^\circ$
$\displaystyle =\sin(90^\circ-5^\circ)+\mathrm{cosec}(90^\circ-5^\circ)$
$\displaystyle =\cos5^\circ+\sec5^\circ$

$\displaystyle \text{(ii) }\mathrm{cosec}\,69^\circ+\cot69^\circ$
$\displaystyle =\mathrm{cosec}(90^\circ-21^\circ)+\cot(90^\circ-21^\circ)$
$\displaystyle =\sec21^\circ+\tan21^\circ$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Prove that:}$
$\displaystyle \text{(i) }\tan1^\circ\tan2^\circ\tan3^\circ\cdots\tan89^\circ=1$
$\displaystyle \text{(ii) }\cos1^\circ\cos2^\circ\cos3^\circ\cdots\cos180^\circ=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) L.H.S.}=\tan1^\circ\tan2^\circ\tan3^\circ\cdots\tan89^\circ$
$\displaystyle =(\tan1^\circ\tan89^\circ)(\tan2^\circ\tan88^\circ)\cdots(\tan44^\circ\tan46^\circ)\tan45^\circ$
$\displaystyle =(\tan1^\circ\cot1^\circ)(\tan2^\circ\cot2^\circ)\cdots(\tan44^\circ\cot44^\circ)\tan45^\circ$
$\displaystyle =1\times1\times\cdots\times1\times1$
$\displaystyle =1=\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}$

$\displaystyle \text{(ii) L.H.S.}=\cos1^\circ\cos2^\circ\cos3^\circ\cdots\cos180^\circ$
$\displaystyle =\cos1^\circ\cos2^\circ\cdots\cos89^\circ\cos90^\circ\cos91^\circ\cdots\cos180^\circ$
$\displaystyle =\cos1^\circ\cos2^\circ\cdots\cos89^\circ\times0\times\cos91^\circ\cdots\cos180^\circ$
$\displaystyle =0=\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }A+B=90^\circ,\text{ prove that:}$
$\displaystyle \sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{\sin^2 B}{\cos^2 A}}=\tan A$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }A+B=90^\circ$
$\displaystyle \therefore B=90^\circ-A$
$\displaystyle \text{L.H.S.}=\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{\sin^2 B}{\cos^2 A}}$
$\displaystyle =\sqrt{\frac{\tan A\tan(90^\circ-A)+\tan A\cot(90^\circ-A)}{\sin A\sec(90^\circ-A)}-\frac{\sin^2(90^\circ-A)}{\cos^2 A}}$
$\displaystyle =\sqrt{\frac{\tan A\cot A+\tan^2 A}{\sin A\,\mathrm{cosec}\,A}-\frac{\cos^2 A}{\cos^2 A}}$
$\displaystyle =\sqrt{1+\tan^2 A-1}$
$\displaystyle =\sqrt{\tan^2 A}$
$\displaystyle =\tan A\quad\text{(since }A\text{ is acute and }\tan A>0\text{)}$
$\displaystyle =\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }A,\ B\text{ and }C\text{ are interior angles of }\triangle ABC,\text{ prove that}$
$\displaystyle \tan\left(\frac{B+C}{2}\right)=\cot\left(\frac{A}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }A+B+C=180^\circ,$
$\displaystyle B+C=180^\circ-A$
$\displaystyle \Rightarrow \frac{B+C}{2}=90^\circ-\frac{A}{2}$
$\displaystyle \therefore \tan\left(\frac{B+C}{2}\right)=\tan\left(90^\circ-\frac{A}{2}\right)$
$\displaystyle =\cot\left(\frac{A}{2}\right)$
$\displaystyle \therefore \tan\left(\frac{B+C}{2}\right)=\cot\left(\frac{A}{2}\right).\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find }\theta\text{ if }\sin(\theta+36^\circ)=\cos\theta\text{ when }\theta+36^\circ$
$\displaystyle \text{is an acute angle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sin(\theta+36^\circ)=\cos\theta$
$\displaystyle \Rightarrow \cos[90^\circ-(\theta+36^\circ)]=\cos\theta$
$\displaystyle \Rightarrow 90^\circ-(\theta+36^\circ)=\theta$
$\displaystyle \Rightarrow 54^\circ-\theta=\theta$
$\displaystyle \Rightarrow 2\theta=54^\circ$
$\displaystyle \therefore \theta=27^\circ$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }\tan2\theta=\cot(\theta+6^\circ),\text{ when }2\theta\text{ and }\theta+6^\circ\text{ are}$
$\displaystyle \text{acute angles, find the value of }\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\tan2\theta=\cot(\theta+6^\circ)$
$\displaystyle \Rightarrow \cot(90^\circ-2\theta)=\cot(\theta+6^\circ)$
$\displaystyle \Rightarrow 90^\circ-2\theta=\theta+6^\circ$
$\displaystyle \Rightarrow 84^\circ=3\theta$
$\displaystyle \therefore \theta=28^\circ$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }A,\ B\text{ and }C\text{ are interior angles of }\triangle ABC,\text{ show that:}$
$\displaystyle \text{(i) }\sin\left(\frac{B+C}{2}\right)=\cos\left(\frac{A}{2}\right)$
$\displaystyle \text{(ii) }\cos\left(\frac{B+C}{2}\right)=\sin\left(\frac{A}{2}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }A+B+C=180^\circ,$
$\displaystyle B+C=180^\circ-A$
$\displaystyle \Rightarrow \frac{B+C}{2}=90^\circ-\frac{A}{2}$
$\displaystyle \text{(i) }\sin\left(\frac{B+C}{2}\right)$
$\displaystyle =\sin\left(90^\circ-\frac{A}{2}\right)$
$\displaystyle =\cos\left(\frac{A}{2}\right)$
$\displaystyle \therefore \sin\left(\frac{B+C}{2}\right)=\cos\left(\frac{A}{2}\right).\text{ Hence proved.}$

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

$\displaystyle \textbf{Question 11: }\text{Find the value of }\theta\text{ for:}$
$\displaystyle \text{(i) }\cos2\theta=\sin4\theta,\text{ where }2\theta,\ 4\theta<90^\circ$
$\displaystyle \text{(ii) }\sin3\theta=\cos(\theta-6^\circ),\text{ where }3\theta,\ (\theta-6^\circ)<90^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\cos2\theta=\sin4\theta$
$\displaystyle \Rightarrow \cos2\theta=\cos(90^\circ-4\theta)$
$\displaystyle \Rightarrow 2\theta=90^\circ-4\theta$
$\displaystyle \Rightarrow 6\theta=90^\circ$
$\displaystyle \therefore \theta=15^\circ$

$\displaystyle \text{(ii) }\sin3\theta=\cos(\theta-6^\circ)$
$\displaystyle \Rightarrow \sin3\theta=\sin[90^\circ-(\theta-6^\circ)]$
$\displaystyle \Rightarrow 3\theta=90^\circ-(\theta-6^\circ)$
$\displaystyle \Rightarrow 3\theta=90^\circ-\theta+6^\circ$
$\displaystyle \Rightarrow 4\theta=96^\circ$
$\displaystyle \therefore \theta=24^\circ$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Prove the following:}$
$\displaystyle \text{(i) }\tan20^\circ\tan35^\circ\tan45^\circ\tan55^\circ\tan70^\circ=1$
$\displaystyle \text{(ii) }\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\,\mathrm{cosec}\,31^\circ=2$
$\displaystyle \text{(iii) }\frac{\cos(90^\circ-A)\sec(90^\circ-A)\tan A}{\mathrm{cosec}(90^\circ-A)\sin(90^\circ-A)\cot(90^\circ-A)}$
$\displaystyle \qquad+\frac{\tan(90^\circ-A)}{\cot A}=2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) L.H.S.}=\tan20^\circ\tan35^\circ\tan45^\circ\tan55^\circ\tan70^\circ$
$\displaystyle =\tan(90^\circ-70^\circ)\tan(90^\circ-55^\circ)\tan45^\circ\tan55^\circ\tan70^\circ$
$\displaystyle =\cot70^\circ\cot55^\circ\tan45^\circ\tan55^\circ\tan70^\circ$
$\displaystyle =\tan45^\circ=1=\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}$

$\displaystyle \text{(ii) L.H.S.}=\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\,\mathrm{cosec}\,31^\circ$
$\displaystyle =\frac{\cos(90^\circ-10^\circ)}{\sin10^\circ}+\cos59^\circ\,\mathrm{cosec}(90^\circ-59^\circ)$
$\displaystyle =\frac{\sin10^\circ}{\sin10^\circ}+\cos59^\circ\sec59^\circ$
$\displaystyle =1+1=2=\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}$

$\displaystyle \text{(iii) L.H.S.}=\frac{\cos(90^\circ-A)\sec(90^\circ-A)\tan A}{\mathrm{cosec}(90^\circ-A)\sin(90^\circ-A)\cot(90^\circ-A)}$
$\displaystyle \qquad+\frac{\tan(90^\circ-A)}{\cot A}$
$\displaystyle =\frac{\sin A\,\mathrm{cosec}\,A\tan A}{\sec A\cos A\tan A}+\frac{\cot A}{\cot A}$
$\displaystyle =\frac{1\times\tan A}{1\times\tan A}+1$
$\displaystyle =1+1=2=\text{R.H.S.}$
$\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{What is the maximum value of:}$
$\displaystyle \text{(i) }\frac{1}{\sec A}\qquad\text{(ii) }\frac{1}{\mathrm{cosec}\,A}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\frac{1}{\sec A}=\cos A$
$\displaystyle \therefore \text{Maximum value}=1$

$\displaystyle \text{(ii) }\frac{1}{\mathrm{cosec}\,A}=\sin A$
$\displaystyle \therefore \text{Maximum value}=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If }\tan A=\frac45,\text{ find the value of }\frac{\cos A-\sin A}{\cos A+\sin A}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{\cos A-\sin A}{\cos A+\sin A}$
$\displaystyle =\frac{1-\tan A}{1+\tan A}$
$\displaystyle =\frac{1-\frac45}{1+\frac45}$
$\displaystyle =\frac{\frac15}{\frac95}=\frac19$
$\displaystyle \therefore \frac{\cos A-\sin A}{\cos A+\sin A}=\frac19$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{If }\tan A=\frac1{\sqrt5},\text{ find the value of}$
$\displaystyle \frac{\mathrm{cosec}^2 A-\sec^2 A}{\mathrm{cosec}^2 A+\sec^2 A}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{\mathrm{cosec}^2 A-\sec^2 A}{\mathrm{cosec}^2 A+\sec^2 A}$
$\displaystyle =\frac{\frac1{\sin^2 A}-\frac1{\cos^2 A}}{\frac1{\sin^2 A}+\frac1{\cos^2 A}}$
$\displaystyle =\frac{\cos^2 A-\sin^2 A}{\cos^2 A+\sin^2 A}$
$\displaystyle =\frac{1-\tan^2 A}{1+\tan^2 A}$
$\displaystyle =\frac{1-\left(\frac1{\sqrt5}\right)^2}{1+\left(\frac1{\sqrt5}\right)^2}$
$\displaystyle =\frac{1-\frac15}{1+\frac15}$
$\displaystyle =\frac{\frac45}{\frac65}=\frac23$
$\displaystyle \therefore \frac{\mathrm{cosec}^2 A-\sec^2 A}{\mathrm{cosec}^2 A+\sec^2 A}=\frac23$
$\displaystyle \\$