Question 14 Marks
If A and B are complementary angles, prove that:
cotB + cosB = secA cosB (1 + sinB)
AnswerSince, A and B are complementary angles, A + B = 90°
cotB + cosB
= cot(90° - A) + cos(90° - A)
= tanA + sinA
$=\frac{\sin A}{\cos A}+\sin A$
$=\frac{\sin A+\sin A \cos A}{\cos A}$
$=\frac{\sin A(1+\cos A)}{\cos A}$
= secA sinA (1 + cosA)
= secA sin(90° - B)(1 + cos(90° - B))
= secA cosB(1 + sinB)
View full question & answer→Question 24 Marks
If $4 \cos2 A – 3 = 0$, Show that:
$\cos 3 A = 4 \cos^3 A – 3 \cos A$
Answer$4 \cos ^2 \mathrm{~A}-3=0$
$\Rightarrow 4 \cos ^2 \mathrm{~A}=3$
$\Rightarrow \cos ^2 \mathrm{~A}=3 / 4$
$\Rightarrow \cos A=\frac{\sqrt{3}}{2}$
$\text { so, } \mathrm{A}=30^{\circ}$
$\text { LHS }=\cos ^3 \mathrm{~A}=\cos 90^{\circ}=0$
$\text { RHS }=4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{~A}$
$=4 \cos ^3 30^{\circ}-3 \cos 30^{\circ}$
$=4\left(\frac{\sqrt{3}}{2}\right)^3-3\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}=0$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 34 Marks
Prove the following identitie:$\frac{(\operatorname{cosec} A-\cot A)^2+1}{\sec A(\operatorname{cosec} A-\cot A)}=2 \cot A$
Answer$\frac{(\operatorname{cosec} A-\cot A)^2+1}{\sec A(\operatorname{cosec} A-\cot A)} $
$ =\frac{(\operatorname{cosec} A-\cot A)^2+\left(\operatorname{cosec}{ }^2 A-\cot ^2 A\right)}{\sec A(\operatorname{cosec} A-\cot A)} $
$ =\frac{(\operatorname{cosec} A-\cot A)^2+(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)}{\sec A(\operatorname{cosec} A-\cot A)}$
$ =\frac{(\operatorname{cosec} A-\cot A)+(\operatorname{cosec} A+\cot A)}{\sec A}$
$ =\frac{2 \operatorname{cosec} A}{\sec A}$
$ =2 \cot A$
View full question & answer→Question 44 Marks
Prove the following identitie:$\frac{1+(\sec A-\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)}=2 \tan A$
Answer$\frac{1+(\sec A-\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)}$
$ =\frac{\left(\sec ^2 A-\tan ^2 A\right)+(\sec A-\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)} $
$ =\frac{(\sec A-\tan A)(\sec A+\tan A)+(\sec A+\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)} $
$ =\frac{(\sec A-\tan A)+(\sec A-\tan A)}{\operatorname{cosec} A} $
$ =\frac{2 \sec A}{\operatorname{cosec} A} $
$=2 \frac{\frac{1}{\cos A}}{\frac{1}{\sin A}} $
$=2 \tan A $
View full question & answer→Question 54 Marks
Prove the following identitie:$\frac{\cos A}{1+\sin A}+\tan A=\sec A$
Answer$\frac{\cos A}{1+\sin A}+\tan A $
$ =\frac{\cos A}{1+\sin A}+\frac{\sin A}{\cos A} $
$ =\frac{\cos ^2 A+\sin A+\sin ^2 A}{(1+\sin ) \cos A} $
$ =\frac{1+\sin A}{(1+\sin A) \cos A} $
$ =\frac{\cos ^3 A+\cos A \sin A-\sin ^2 A}{\cos ^2 A-\sin A \cos A} $
$=\frac{1}{\cos A} $
$=\sec A$
View full question & answer→Question 64 Marks
Prove the following identitie:$\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}=1+\tan A+\cot A$
Answer$\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A} $
$=\frac{\frac{1}{\tan A}}{1-\tan A}+\frac{\tan A}{1-\frac{1}{\tan A}} $
$ =\frac{1}{\tan A(1-\tan A)}+\frac{\tan ^2 A}{\tan A-1} $
$=\frac{1-\tan ^3 A}{\tan A(1-\tan A)} $
$ =\frac{(1-\tan A)\left(1+\tan A+\tan ^2 A\right)}{\tan A(1-\tan A)} $
$=\frac{1+\tan A+\tan ^2 A}{\tan A} $
$ =\cot A +1+\tan A$
View full question & answer→Question 74 Marks
Prove the following identitie:$\frac{1-\cos A}{\sin A}+\frac{\sin A}{1-\cos A}=2 \operatorname{cosec} A$
Answer$\frac{1-\cos A}{\sin A}+\frac{\sin A}{1-\cos A} $
$ =\frac{(1-\cos A)^2+\sin ^2 A}{\sin A(1-\cos A)} $
$ =\frac{1+\cos ^2 A-2 \cos A+\sin ^2 A}{\sin A(1-\cos A)} $
$ \frac{2-2 \cos A}{\sin A(1-\cos A)} $
$\frac{2(1-\cos A)}{\sin A(1-\cos A)} $
$=2 \operatorname{cosec} A$
View full question & answer→Question 84 Marks
Prove that$\sqrt{\sec ^2 A+\operatorname{cosec} c^2 A}=\tan A+\cot A$
Answer$\text { LHS }=\sqrt{\sec ^2 A+\operatorname{cosec}{ }^2 A} $
$=\sqrt{\frac{1}{\cos ^2 A}+\frac{1}{\sin ^2 A}} $
$ =\sqrt{\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 A \cos ^2 A}} $
$ =\sqrt{\frac{1}{\sin ^2 A \cos ^2 A}} $
$ =\frac{1}{\sin A \cos A}$
$\text { RHS }=\tan A+\cot A $
$ =\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} $
$ =\frac{\sin ^2 A+\cos ^2 A}{\sin A \cos A} $
$ =\frac{1}{\sin A \cos A} $
$ \text { LHS }=\text { RHS }$
View full question & answer→Question 94 Marks
Without using trigonometrical tables, evaluate:$\operatorname{cosec} 257^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \operatorname{cosec} 46^{\circ}-\sqrt{2} \cos 45^{\circ}-\tan ^2 60^{\circ}$
Answer$\cos e c^2 57^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \cos e c 46^{\circ}-\sqrt{2} \cos 45^{\circ}-\tan ^2 60^{\circ} $
$ =\cos e c^2\left(90^{\circ}-33^{\circ}\right)^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \cos e c\left(90^{\circ}-44^{\circ}\right)-\sqrt{2} \cos 45^{\circ}-\tan ^2 60^{\circ}$
$=\sec ^2 33^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \sec 44^{\circ}-\sqrt{2} \cos 45^{\circ}-\tan ^2 60 $
$ =1+1-\sqrt{2} \cos 45^{\circ}-\tan ^2 60$
$ =1+1-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)-(\sqrt{3})^2 $
$=2-1-3 $
$ =-2$
View full question & answer→Question 104 Marks
Evaluate
$\sin ^2 34^{\circ}+\sin ^2 56^{\circ}+2 \tan 18^{\circ} \tan 72^{\circ}-\cot ^2 30^{\circ}$
Answer$\sin ^2 34^{\circ}+\sin ^2 56^{\circ}+2 \tan 18^{\circ} \tan 72^{\circ}-\cot ^2 30^{\circ}$
$=\sin ^2 34^{\circ}+\sin ^2\left(90^{\circ}-34^{\circ}\right)+23 \tan 18^{\circ} \tan \left(90^{\circ}-72^{\circ}\right)-\cot ^2 30^{\circ}$
$=\sin ^2 34^{\circ}+\cos ^2 34^{\circ}+2 \tan 18^{\circ} \cot ^2 18^{\circ}-\cot ^2 30^{\circ}$
$=\left(\sin ^2 34^{\circ}+\cos ^2 34^{\circ}\right)+2 \tan 18^{\circ} \times \frac{1}{\tan 18^{\circ}}-\cot ^2 30^{\circ}$
$=1+2 \times 1-(\sqrt{3})^2$
$=1+2-3$
$=3-3$
$=0$
View full question & answer→Question 114 Marks
If $\frac{\cos A}{\cos B}=m$ and $\frac{\cos A}{\sin B}= n$ show that:
$\left(m^2+n^2\right) \cos ^2 B=n^2$.
Answer$\text { LHS }=\left(m^2+n^2\right) \cos ^2 B$
$ =\left(\frac{\cos ^2 A}{\cos ^2 B}+\frac{\cos ^2 A}{\sin ^2 B}\right) \cos ^2 B$
$ =\left(\frac{\cos ^2 A \sin ^2 B+\cos ^2 A \cos ^2 B}{\cos ^2 B \sin ^2 B}\right) \cos ^2 B$
$=\left(\frac{\cos ^2 A \sin ^2 B+\cos ^2 A \cos ^2 B}{\sin ^2 B}\right)$
$ =\frac{\cos ^2 A\left(\sin ^2 B+\cos ^2 B\right)}{\sin ^2 B} $
$=\frac{\cos ^2 A}{\sin ^2 B} $
$ =n^2$
Hence, $(m^2 + n^2) \cos^2B = n^2.$
View full question & answer→Question 124 Marks
If $\sin A+\cos A=m$ and $\sec A+\operatorname{cosec} A=n$, show that: $n\left(m^2-1\right)=2 m$
AnswerGiven:
sin A + cos A = m
and
sec A + cosec A = n
Consider $L.H.S =n (m^2 - 1)$
$=(\sec A+\cos e c A)\left((\sin A+\cos A)^2-1\right)$
$=\left(\frac{1}{\cos A}+\frac{1}{\sin A}\right)\left(\sin ^2 A+\cos ^2 A+2 \sin A \cos A-1\right)$
$=\left(\frac{\cos A+\sin A}{\sin A \cos A}\right)(1+2 \sin A \cos A-1)$
$=\frac{\cos A+\sin A}{\sin A \cos A}(2 \sin A \cos A)$
= 2 (sinA + cosA)
= 2m = RHS
View full question & answer→Question 134 Marks
Prove.$\frac{1}{\cos A+\sin A-1}+\frac{1}{\cos A+\sin A+1}=\operatorname{cosec} A+\sec A$
Answer$\text { LHS }=\frac{1}{\cos A+\sin A-1}+\frac{1}{\cos A+\sin A+1}$
$=\frac{\cos A+\sin A+1+\cos A+\sin A-1}{(\cos A+\sin A)^2-1} $
$=\frac{2(\cos A+\sin A)}{\cos ^2 A+\sin ^2 A+2 \cos A \sin A-1} $
$ =\frac{2(\cos A+\sin A)}{1+2 \cos A \sin A}=\frac{\cos A+\sin A}{\cos A \sin A} $
$ =\frac{\cos A}{\cos A \sin A}+\frac{\sin A}{\cos A \sin A} $
$ =\frac{1}{\sin A}+\frac{1}{\cos A} $
$=\operatorname{cosec} A+\sec A=\text { RHS }$
View full question & answer→Question 144 Marks
Prove.$\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin 3 A}=2$
Answer$\text { LHS }=\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin 3 A}$
$ =\frac{\left(\cos ^3 A+\sin ^3 A\right)(\cos A-\sin A)+\left(\cos ^3 A-\sin ^3 A\right)(\cos A+\sin A)}{\cos ^2 A-\sin ^2 A} $
$ =\left(\cos ^4 A -\cos ^3 A \sin A +\sin ^3 A \cos A -\sin ^4 A +\cos ^4 A +\cos ^3 \prime A \sin A-\sin ^{\wedge} 3 A \cos A-\sin ^{\wedge} 4 A\right) /\left(\cos ^{\wedge} 2 A-\sin ^{\wedge} 2 A\right)$
$ =\frac{2\left(\cos ^4 A-\sin ^4 A\right)}{\cos ^2 A-\sin ^2 A} $
$=\frac{2\left(\cos ^2 A+\sin ^2 A\right)\left(\cos ^2 A-\sin ^2 A\right)}{\cos ^2 A-\sin ^2 A} $
$=2\left(\cos ^2 A +\sin ^2 A \right)$
$ =2=\text { RHS } \quad\left(\because \cos ^2 A +\sin ^2 A =1\right) $
View full question & answer→Question 154 Marks
Prove.$\frac{\sin A+\cos A}{\cos A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2 \sin ^2 A-1}$
Answer$\text { LHS }=\frac{\sin A+\cos A}{\cos A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A} $
$=\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A+\cos A)(\sin A-\cos A)} $
$ \frac{\sin ^2 A+\cos ^2 A+2 \sin A \cos A+\sin ^2 A+\cos ^2 A-2 \sin A \cos A}{\sin ^2 A-\cos ^2 A} $
$ \frac{2\left(\sin ^2 A+\cos ^2 A\right)}{\sin ^2 A-\cos ^2 A} $
$ =\frac{2}{\sin ^2 A-\cos ^2 A}\left(\sin ^2 A+\cos ^2 A=1\right)$
$=\frac{2}{\sin ^2 A-\cos ^2 A}=\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)} $
$ =\frac{2}{2 \sin ^2 A-1}=\text { RHS }$
View full question & answer→Question 164 Marks
Prove.
$(1 + \cot A - \cos ec A)(1+ \tan A + \sec A) = 2$
Answer$\text { LHS }=(1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A)$
$=\left(1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}-\frac{1}{\cos A}\right)$
$=\left(\frac{\sin A+\cos A-1}{\sin A}\right)\left(\frac{\cos A+\sin A+1}{\cos A}\right)$
$=\frac{(\sin A+\cos A-1)(\sin A+\cos A+1)}{\sin A \cos A}$
$=\frac{(\sin A+\cos A)^2-(1)^2}{\sin A \cos A}$
$=\frac{\sin ^2 A+\cos ^2 A+2 \sin A \cos A-1}{\sin A \cos A}$
$=\frac{1+2 \sin A \cos A-1}{\sin A \cos A}$
$=\frac{2 \sin A \cos A}{\sin A \cos A}=2=\text { RHS }$
View full question & answer→Question 174 Marks
Prove the following identitie:$\frac{(\operatorname{cosec} A-\cot A)^2+1}{\sec A(\operatorname{cosec} A-\cot A)}=2 \cot A$
Answer$\frac{(\operatorname{cosec} A-\cot A)^2+1}{\sec A(\operatorname{cosec} A-\cot A)}$
$=\frac{(\operatorname{cosec} A-\cot A)^2+\left(\operatorname{cosec}{ }^2 A-\cot ^2 A\right)}{\sec A(\operatorname{cosec} A-\cot A)}$
$=\frac{(\operatorname{cosec} A-\cot A)^2+(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)}{\sec A(\operatorname{cosec} A-\cot A)}$
$=\frac{(\operatorname{cosec} A-\cot A)+(\operatorname{cosec} A+\cot A)}{\sec A}$
$=\frac{2 \operatorname{cosec} A}{\sec A}$
$=2 \cot A$
View full question & answer→Question 184 Marks
Prove that$\sqrt{\sec ^2 A+\operatorname{cosec} c^2 A}=\tan A+\cot A$
Answer$\text { LHS }=\sqrt{\sec ^2 A+\operatorname{cosec}{ }^2 A}$
$=\sqrt{\frac{1}{\cos ^2 A}+\frac{1}{\sin ^2 A}}$
$=\sqrt{\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 A \cos ^2 A}}$
$=\sqrt{\frac{1}{\sin ^2 A \cos ^2 A}}$
$=\frac{1}{\sin A \cos A}$
$\text { RHS }=\tan A+\cot A$
$=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$
$=\frac{\sin ^2 A+\cos ^2 A}{\sin A \cos A}$
$=\frac{1}{\sin A \cos A}$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 194 Marks
Prove.$\frac{1}{\cos A+\sin A-1}+\frac{1}{\cos A+\sin A+1}=\operatorname{cosec} A+\sec A$
Answer$\text { LHS }=\frac{1}{\cos A+\sin A-1}+\frac{1}{\cos A+\sin A+1}$
$=\frac{\cos A+\sin A+1+\cos A+\sin A-1}{(\cos A+\sin A)^2-1}$
$=\frac{2(\cos A+\sin A)}{\cos ^2 A+\sin ^2 A+2 \cos A \sin A-1}$
$=\frac{2(\cos A+\sin A)}{1+2 \cos A \sin A}=\frac{\cos A+\sin A}{\cos A \sin A}$
$=\frac{\cos A}{\cos A \sin A}+\frac{\sin A}{\cos A \sin A}$
$=\frac{1}{\sin A}+\frac{1}{\cos A}$
$=\operatorname{cosec} A+\sec A=\text { RHS }$
View full question & answer→Question 204 Marks
Prove.$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
Answer$\text { LHS }=(\cos e c A-\sin A)(\sec A-\cos A)$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$
$=\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right)$
$=\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)$
$=\sin A \cos A$
$\text { RHS }=\frac{1}{\tan A+\cot A}$
$=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}$
$=\frac{\sin A \cos A}{\sin A+\cos 2} A$
$=\sin A \cos A$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 214 Marks
Prove.$\frac{\tan A}{1-\cot A}+\frac{\cot }{1-\tan A}=\sec A \operatorname{cosec} A+1$
Answer$\text { LHS }=\frac{\tan A}{1-\cot A}+\frac{\cot }{1-\tan A}$
$=\frac{\tan A}{1-\frac{1}{\tan A}}+\frac{\frac{1}{\tan A}}{1-\tan A}$
$=\frac{\tan ^2 A}{\tan A-1}+\frac{1}{\tan A(1-\tan A)}$
$=\frac{\tan ^3 A-1}{\tan A(1-\tan A)}$
$=\frac{\left(\tan ^2-1\right)(\tan A+1+\tan A)}{\tan A(\tan A-1)}$
$=\frac{\sec ^2 A+\tan A}{\tan A}$
$=\frac{\frac{1}{\cos ^2 A}}{\frac{\sin A}{\cos ^2}+1}=\frac{1}{\sin A \cos A}+1$
$=\sec A \operatorname{cosec} A+1=\text { RHS }$
View full question & answer→Question 224 Marks
Prove the following trigonometric identities.
$sec A\ (1 − \sin A) (sec\ A + \tan A) = 1$
AnswerWe have to prove $sec\ A\ (1 − \sin A)(sec\ A + \tan A) = 1$
We know that $sec^2\ A − \tan^2 A − 1$
So$,$
$sec\ A(1 − \sin A)(sec\ A + \tan A) = {sec \ A(1 − \sin A)}(sec A + \tan A)$
$= (sec\ A − sec\ A \sin A)(sec\ A + \tan A)$
$=\left(\sec\ A-\frac{1}{\cos A} \sin A\right)(\sec\ A+\tan A) \ldots\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$
$=\left(\sec\ A-\frac{\sin A}{\cos A}\right)(\sec\ A+\tan A) \ldots\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$
$=(\sec A-\tan A)(\sec A+\tan A)$
$=\sec ^2 A-\tan ^2 A$
$=1=\text { R.H.S. }\left(\because \sec ^2 \theta=1 \tan ^2 \theta\right)$
View full question & answer→Question 234 Marks
Prove.$\frac{\sin A+\cos A}{\cos A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2 \sin ^2 A-1}$
Answer$\begin{aligned} & \text { LHS }=\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A} \\ & =\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)} \\ & =\frac{\sin ^2 A+\cos ^2 A+2 \sin A \cos A+\sin ^2 A+\cos ^2 A-2 \sin A \cos A}{\sin ^2 A-\cos ^2 A} \\ & =\frac{2\left(\sin ^2 A+\cos ^2 A\right)}{\sin ^2 A-\cos ^2 A} \\ & \left.=\frac{2}{\sin ^2 A-\cos ^2 A} \cdots \sin ^2 A+\cos ^2 A=1\right] \\ & =\frac{2}{\sin ^2 A-\cos ^2 A} \\ & =\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)} \\ & =\frac{2}{2 \sin ^2 A-1} \\ & =\text { RHS }\end{aligned}$
View full question & answer→Question 244 Marks
Prove.$\frac{1}{\sec A+\tan A}=\sec A-\tan A$
Answer$
\begin{aligned}
& \text { LHS }=\frac{1}{\sec A +\tan A } \\
& =\frac{1}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}} \\
& =\frac{1}{\frac{1+\sin A }{\cos A }} \\
& =\frac{\cos A }{1+\sin A } \times \frac{1-\sin A }{1+\sin A } \\
& =\frac{\cos A(1-\sin A)}{(1)^2-\sin ^2 A} \\
& =\frac{\cos A(1-\sin A)}{\cos ^2 A} \\
& =\frac{1}{\cos A}-\frac{\sin A}{\cos A} \\
& =\sec A -\tan A \\
& \text { LHS }=\text { RHS }
\end{aligned}
$Hence proved.q
View full question & answer→