Question 11 Mark
Without using trignometric tables, prove that:
$\text{cosec }^272^\circ-\tan^218^\circ=1$
Answer$\text{L.H.S.}=\text{cosec }^272^\circ-\tan^218^\circ$
$=\text{cosec}^2\big(90^\circ-18^\circ\big)-\tan18^\circ$
$=\sec^218^\circ-\tan^218^\circ$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 21 Mark
Without using trignometric tables, prove that:
$\big(\sin^265^\circ+\cos^225^\circ\big)\big(\sin65^\circ-\cos25^\circ\big)=0$
Answer$\text{L.H.S.}=\big(\sin^265^\circ+\cos^225^\circ\big)\big(\sin65^\circ-\cos25^\circ\big)$
$=\sin^265^\circ-\cos^225^\circ$
$=\sin^2\big(90^\circ-25^\circ\big)-\cos^225^\circ$
$=\cos^225^\circ-\cos25^\circ$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 31 Mark
Without using trigomentric tables, prove that:
$\sin35^\circ\sin55^\circ+\cos35^\circ\cos55^\circ=0$
Answer$\text{L.H.S}=\sin35^\circ\sin55^\circ+\cos35^\circ\cos55^\circ$
$=\sin35^\circ\cos\big(90^\circ-55^\circ\big)-\cos35^\circ\sin\big(90^\circ-55^\circ\big)$
$=\sin35^\circ\cos35^\circ-\cos35^\circ\sin35^\circ$
$=0$
$=\text{R.H.S.}$
View full question & answer→Question 41 Mark
Without using trigomentric tables, prove that:
$\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ=1$
Answer$\text{L.H.S}=\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$
$=\cot\big(90^\circ+48^\circ\big)\cot\big(90^\circ-23^\circ\big)-\tan42^\circ\tan67^\circ$
$=\cot42^\circ\cot67^\circ\tan42^\circ\tan67^\circ$
$=\frac{1}{\tan42^\circ}\times\frac{1}{\tan67^\circ}\times\tan42^\circ\times\tan67^\circ$
$=1$
$=\text{R.H.S.}$
View full question & answer→Question 51 Mark
Without using trignometric tables, prove that:
$\text{cosec}80^\circ-\sec10^\circ=0$
Answer$\text{L.H.S.}=\text{cosec}80^\circ-\sec10^\circ$
$=\text{cosec}\big(90^\circ-10^\circ\big)-\sec10^\circ$
$=\sec10^\circ-\sec10^\circ$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 61 Mark
Express the following in terms of T-ratio of angles lying between 0° and 45°.
$\sec78^\circ+\text{cosec }56^\circ$
Answer$\sec78^\circ+\text{cosec }56^\circ$
$=\sec(90^\circ-65^\circ)+\cot(90^\circ-49^\circ)$
$=\text{cosec }12^\circ+\sec34^\circ$
View full question & answer→Question 71 Mark
Without using trignometric tables, prove that:
$\cos81^\circ-\sin9^\circ=0$
Answer$\text{L.H.S.}=\cos81^\circ-\sin9^\circ$
$=\cos\big(90^\circ-9^\circ\big)-\sin9^\circ$
$=\sin9^\circ-\sin9^\circ$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 81 Mark
Express the following in terms of T-ratio of angles lying between 0° and 45°.
$\cot65^\circ+\tan49^\circ$
Answer$\cot65^\circ+\tan49^\circ$
$=\tan(90^\circ-65^\circ)+\cot(90^\circ-49^\circ)$
$=\tan25^\circ+\cot41^\circ$
View full question & answer→Question 91 Mark
Without using trigomentric tables, prove that:
$\sec70^\circ\sin20^\circ+\cos20^\circ\text{cosec}70^\circ=2$
Answer$\text{L.H.S}=\sec70^\circ\sin20^\circ+\cos20^\circ\text{cosec}70^\circ$
$=\sec\big(90^\circ-20^\circ\big)\sin20^\circ+\cos20^\circ\text{cosec}\big(90^\circ-20^\circ\big)$
$=\text{cosec}20^\circ.\frac{1}{\text{cosec}20^\circ}+\frac{1}{\sec20^\circ}.\sec20^\circ$
$=1+1$
$=2$
$=\text{R.H.S.}$
View full question & answer→Question 101 Mark
Without using trigomentric tables, prove that:
$\big(\sin72^\circ+\cos18^\circ\big)\big(\sin72^\circ-\cos18^\circ\big)=0$
Answer$\text{L.H.S}=\big(\sin72^\circ+\cos18^\circ\big)\big(\sin72^\circ-\cos18^\circ\big)$
$=\big(\sin72^\circ+\cos18^\circ\big)\big[\cos\big(90^\circ-72^\circ\big)-\cos18^\circ\big]$
$=\big(\sin72^\circ+\cos18^\circ\big)\big(\cos18^\circ-\cos18^\circ\big)$
$=\big(\sin72^\circ+\cos18^\circ\big)(0)$
$=0$
$=\text{R.H.S.}$
View full question & answer→Question 111 Mark
Without using trigonometric tables, evaluate:
$\frac{\text{cosec}42^\circ}{\sec48^\circ}$
Answer$\frac{\text{cosec}42^\circ}{\sec48^\circ}$
$=\frac{\text{cosec}(90^\circ-48^\circ)}{\sec48^\circ}$
$=\frac{\sec48^\circ}{\sec48^\circ}$ $\big[\because\sec(90-\theta)=\text{cosec }\theta\big]$
$=1$
View full question & answer→Question 121 Mark
Without using trignometric tables, evaluate:
$\frac{\sec11^\circ}{\text{cosec}79^\circ}$
Answer$\frac{\sec11^\circ}{\text{cosec}79^\circ}$
$=\frac{\sec(90^\circ-79^\circ)}{\text{cosec}79^\circ}$
$=\frac{\text{cosec}79^\circ}{\text{cosec}79^\circ}$ $[\because\ \sec(90-\theta)=\text{cosec }\theta]$
$=1$
View full question & answer→Question 131 Mark
Without using trigomentric tables, prove that:
$\sin53^\circ\cos37^\circ+\cos53^\circ\sin37^\circ=1$
Answer$\text{L.H.S}=\sin53^\circ\cos37^\circ+\cos53^\circ\sin37^\circ$
$=\sin\big(90^\circ-37^\circ\big)\cos37^\circ+\cos\big(90^\circ-37^\circ\big)\sin37^\circ$
$=\cos37^\circ\cos37^\circ+\sin37^\circ\sin37^\circ$
$=\cos^237^\circ+\sin^237$
$=1$
$=\text{R.H.S.}$
View full question & answer→Question 141 Mark
Without using trignometric tables, prove that:
$\sin^248^\circ+\sin^242^\circ=1$
Answer$\text{L.H.S.}=\sin^248^\circ+\sin^242^\circ$
$=\sin^2\big(90^\circ-42^\circ\big)+\sin^242^\circ$
$=\cos^242^\circ+\sin^242^\circ$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 151 Mark
Without using trignometric tables, prove that:
$\cos^257^\circ+\sin^233^\circ=0$
Answer$\text{L.H.S.}=\cos^257^\circ-\sin^233^\circ$
$=\cos^2\big(90^\circ-33^\circ\big)-\sin^233^\circ$
$=\sin^233^\circ-\sin^233^\circ$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 161 Mark
Without using trigonometric tables, evaluate:
$\frac{\cos35^\circ}{\sin55^\circ}$
Answer$\frac{\cos35^\circ}{\sin55^\circ}$
$=\frac{\cos(90^\circ-55^\circ)}{\sin55^\circ}$
$=\frac{\sin55^\circ}{\sin55^\circ}$ $\big[\because\sin(90-\theta)=\cos\theta\big]$
$=1$
View full question & answer→Question 171 Mark
Without using trignometric tables, prove that:
$\tan^266^\circ+\cot^224^\circ=0$
Answer$\text{L.H.S.}=\tan^266^\circ+\cot^224^\circ$
$=\tan^2\big(90^\circ-24^\circ\big)-\cot^224^\circ$
$=\cot^224^\circ-\cot^224^\circ$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 181 Mark
Without using trignometric tables, prove that:
$\tan71^\circ-\cot19^\circ=0$
Answer$\text{L.H.S.}=\tan71^\circ-\cot19^\circ$
$=\tan\big(90^\circ-19^\circ\big)-\cot19^\circ$
$=\cot19^\circ-\cot19^\circ$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 191 Mark
Without using trignometric tables, evaluate:
$\frac{\sin16^\circ}{\cos74^\circ}$
Answer$\frac{\sin16^\circ}{\cos74^\circ}$
$=\frac{\sin(90^\circ-74^\circ)}{\cos74^\circ}$
$=\frac{\cos74^\circ}{\cos74^\circ}$ $[\therefore\ \sin(90-\theta)=\cos\theta]$
$=1$
View full question & answer→Question 201 Mark
Without using trignometric tables, evaluate:
$\frac{\cot38^\circ}{\tan52^\circ}$
Answer$\frac{\cot38^\circ}{\tan52^\circ}$
$=\frac{\cot(90^\circ-52^\circ)}{\tan52^\circ}$
$=\frac{\tan52^\circ}{\tan52^\circ}$ $\big[\therefore\ \tan(90-\theta)=\cot\theta\big]$
$=1$
View full question & answer→Question 211 Mark
Without using trigomentric tables, prove that:
$\cos54^\circ\cos36^\circ+\sin54^\circ\sin36^\circ=0$
Answer$\text{L.H.S}=\cos54^\circ\cos36^\circ+\sin54^\circ\sin36^\circ$
$=\cos\big(90^\circ-36^\circ\big)\cos36^\circ+\sin\big(90^\circ-36^\circ\big)\sin36^\circ$
$=\sin36^\circ\cos36^\circ+\cos36^\circ\sin36^\circ$
$=0$
$=\text{R.H.S.}$
View full question & answer→Question 221 Mark
Without using trignometric tables, prove that:
$\cos^275^\circ+\cos^215^\circ=1$
Answer$\text{L.H.S.}=\cos^275^\circ+\cos^215^\circ$
$=\cos^2\big(90^\circ-15^\circ\big)-\cos^215^\circ$
$=\sin^215^\circ-\cos^215^\circ$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 231 Mark
Express the following in terms of T-ratio of angles lying between 0° and 45°.
$\text{cosec}54^\circ+\sin72^\circ$
Answer$\text{cosec}54^\circ+\sin72^\circ$
$=\sec(90^\circ-54^\circ)+\cos(90^\circ-72^\circ)$
$=\sec36^\circ+\cos18^\circ$
View full question & answer→Question 241 Mark
Express the following in terms of T-ratio of angles lying between 0° and 45°.
$\sin67^\circ+\cos75^\circ$
Answer$\sin67^\circ+\cos75^\circ$
$=\cos(90^\circ-67^\circ)+\sin(90^\circ-75^\circ)$
$=\cos(23^\circ+\sin15^\circ)$
View full question & answer→Question 251 Mark
Without using trignometric tables, evaluate:
$\frac{\tan27^\circ}{\cot63^\circ}$
Answer$\frac{\tan27^\circ}{\cot63^\circ}$
$=\frac{\tan(90^\circ-63^\circ)}{\cot63^\circ}$
$=\frac{\cot63^\circ}{\cot63^\circ}$ $[\therefore\ \tan(90-\theta)=\cot\theta]$
$=1$
View full question & answer→