Download App

Mathematics [3 marks sum]

Trigonometrical Identities · ICSE STD 10 (ICSE - English Medium). 122 questions.

Questions · Page 3 of 3

[3 marks sum]

Question 1013 Marks
Prove. $(1 + \tan A \tan B)^2 + ( \tan A - \tan B)^2 = \sec ^2A \sec ^2B$
Answer
$\text{LHS} =(1 + \tan A \ \tan B)^2 + ( \tan A - \tan B)^2$
$= 1 + \ \tan ^2A \ \tan ^2B + 2 \tan A \ \tan B+\ \tan ^2A + \ \tan ^2B - 2 \tan A \ \tan B$
$= 1 + \ \tan ^2A + \ \tan ^2B + \ \tan ^2A \ \tan ^2B$
$= \sec^2A + \ \tan ^2B (1 + \ \tan ^2A)$
$= \sec^2A + \ \tan ^2B \ sec^2A$
$= \sec^2A (1 + \ \tan ^2B)$
$= \sec^2A sec^2B = \text{RHS}$
View full question & answer→
Question 1023 Marks
Prove.$\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0$
Answer
$\text { LHS }=\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}$
$=\frac{\sin ^2 A-\sin ^2 B+\cos ^2 A-\cos ^2 B}{(\cos A+\cos B)(\sin A+\sin B)}$
$=\frac{\left(\sin ^2 A+\cos ^2 A\right)-\left(\sin ^2 B+\cos ^2 B\right)}{(\cos A+\cos B)(\sin A+\sin B)}$
$\frac{1-1}{(\cos A+\cos B)(\sin A+\sin B)}$
$=0=\text { RHS }$
View full question & answer→
Question 1033 Marks
Prove.$\left(\tan A+\frac{1}{\cos A}\right)^2+\left(\tan A-\frac{1}{\cos A}\right)^2=2\left(\frac{1+\sin ^2 A}{1-\sin ^2 A}\right)$
Answer
$\text { LHS }=\left(\tan A+\frac{1}{\cos A}\right)^2+\left(\tan A-\frac{1}{\cos A}\right)^2$
$=\left(\frac{\sin A+1}{\cos A}\right)^2+\left(\frac{\sin A-1}{\cos A}\right)^2$
$=\frac{\sin ^2 A+1+2 \sin A+\sin ^2 A+1-2 \sin A}{\cos ^2 A}$
$=\frac{2+2 \sin ^2 A}{\cos ^2 A}$
$=2\left(\frac{1+\sin ^2 A}{1-\sin ^2 A}\right)=\text { RHS }$
View full question & answer→
Question 1043 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
$\text { LHS }=\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin 3 A}$
$=\frac{\left(\cos ^3 A+\sin ^3 A\right)(\cos A-\sin A)+\left(\cos ^3 A-\sin ^3 A\right)(\cos A+\sin A)}{\cos ^2 A-\sin ^2 A}$
$=\left(\cos ^4 A -\cos ^3 A \sin A +\sin ^3 A \cos A -\sin ^4 A +\cos ^4 A +\cos ^3 1 A \sin A-\sin^3 A \cos A-\sin^4 A\right) /\left(\cos^2 A-\sin^2A\right)$
$=\frac{2\left(\cos ^4 A-\sin ^4 A\right)}{\cos ^2 A-\sin ^2 A}$
$=\frac{2\left(\cos ^2 A+\sin ^2 A\right)\left(\cos ^2 A-\sin ^2 A\right)}{\cos ^2 A-\sin ^2 A}$
$=2\left(\cos ^2 A+\sin ^2 A\right)$
$=2=\text { RHS }\left(\because \cos ^2 A+\sin ^2 A=1\right)$
View full question & answer→
Question 1053 Marks
Prove.$\frac{\cos \theta \cot \theta}{1+\sin \theta}=\operatorname{cosec} \theta-1$
Answer
$\text { LHS }=\frac{\cos \theta \cot \theta}{1+\sin \theta}$
$=\frac{\cos \theta \cot \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{\cos \theta \cot \theta(1-\sin \theta)}{1-\sin \theta}$
$=\frac{\cos \theta \frac{\cos \theta}{\sin \theta}(1-\sin \theta)}{\cos ^2 \theta}$
$=\frac{1-\sin \theta}{\sin \theta}$
$=\frac{1}{\sin \theta}-1$
$=\operatorname{cosec} \theta-1$
View full question & answer→
Question 1063 Marks
Prove:
$\frac{1+\sin A}{\operatorname{cosec} A-\cot A}-\frac{1-\sin A}{\operatorname{cosec} A+\cot A}=2(1+\cot A)$
Answer
$
\begin{aligned}
& \text { LHS }=\frac{1+\sin A}{\operatorname{cosec} A-\cot A}-\frac{1-\sin A}{\operatorname{cosec} A+\cot A} \\
& =\frac{(1+\sin A)(\operatorname{cosec} A+\cot A)-(1-\sin A)(\operatorname{cosec} A-\cot A)}{(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)} \\
& =\frac{\operatorname{cosec} A+\cot A+\sin A \operatorname{cosec} A+\sin A \cot A-\operatorname{cosec} A+\cot A+\sin A \operatorname{cosec} A-\sin A \cos A}{\operatorname{cosec}^2 A-\cot ^2 A} \\
& =2 \cot A+2 \sin A \operatorname{cosec} A \\
& =2 \cot A+2 \frac{1}{\operatorname{cosec} A} \times \operatorname{cosec} A \\
& =2(\cot A+1)
\end{aligned}
$Hence proved.
View full question & answer→
Question 1073 Marks
Prove.$\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}=\frac{1+\cos A}{\sin A}$
Answer
$\text { LHS }=\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}$
$=\frac{\cot A+\operatorname{cosec} A-\left(\operatorname{cosec}{ }^2 A-\cot ^2 A\right)}{\cot A-\operatorname{cosec} A+1}\left(\operatorname{cosec}^2 A-\cot ^2 A=1\right)$
$=\frac{\cot A+\operatorname{cosec} A-((\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A))}{\cot A-\operatorname{cosec} A+1}$
$=\frac{\cot A+\operatorname{cosec} A(1-\operatorname{cosec} A+\cot A)}{\cot A-\operatorname{cosec} A+1}$
$=\cot A +\operatorname{cosec} A$
$=\frac{\cos A}{\sin A}+\frac{1}{\sin A}$
$=\frac{1+\cos A}{\sin A}$
View full question & answer→
Question 1083 Marks
Prove.
$(1 + \cot A - \operatorname{cosec} A)(1+ \tan A + \sec A) = 2$
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 1093 Marks
Prove.$\frac{\sin A \tan A}{1-\cos A}=1+\sec A$
Answer
$\text { LHS }=\frac{\sin A \tan A}{1-\cos A}$
$=\frac{\sin A \tan A}{1-\cos A} \times \frac{1+\cos A}{1+\cos A}$
$=\frac{\sin A \tan A(1+\cos A)}{1-\cos ^2 A}$
$=\frac{\sin A \frac{\sin A}{\cos A}(1+\cos A)}{\sin ^2 A}$
$=\frac{1+\cos A}{\cos A}$
$=\frac{1}{\cos A}+\frac{\cos A}{\cos A}$
$=\sec A+1$
View full question & answer→
Question 1103 Marks
Prove.$\frac{\sin A}{1+\cos A}=\operatorname{cosec} A-\cot A$
Answer
$\text { LHS }=\frac{\sin A}{1+\cos A}$
$=\frac{\sin A}{1+\cos A} \times \frac{1-\cos A}{1-\cos A}$
$=\frac{\sin A(1-\cos A)}{1-\cos ^2 A}$
$=\frac{1-\cos A}{\sin A}$
$=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$
$=\operatorname{cosec} A-\cot A=\text { RHS }$
View full question & answer→
Question 1113 Marks
Prove:
$\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}=\tan \theta$
Answer
$\begin{aligned} & \text { LHS }=\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta} \\ & =\frac{\sin \theta\left(1-2 \sin ^2 \theta\right)}{\cos \theta\left(2 \cos ^2 \theta-1\right)} \\ & =\frac{\sin \theta\left(\sin ^2 \theta+\cos ^2 \theta-2 \sin ^2 \theta\right)}{\cos \theta\left(2 \cos ^2 \theta-\sin ^2 \theta-\cos ^2 \theta\right)} \\ & =\frac{\sin \theta\left(\cos ^2 \theta-\sin ^2 \theta\right)}{\cos \theta\left(\cos ^2 \theta-\sin ^2 \theta\right)} \\ & =\frac{\sin \theta}{\cos \theta} \\ & =\tan \theta=\text { RHS }\end{aligned}$
View full question & answer→
Question 1123 Marks
Prove.$\tan ^2 A-\tan ^2 B=\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}$
Answer
$\text { LHS }=\tan ^2 A-\tan ^2 B$
$=\frac{\sin ^2 A}{\cos ^2 A}-\frac{\sin ^2 B}{\cos ^2 B}$
$=\frac{\sin ^2 A \cdot \cos ^2 B-\sin ^2 B \cdot \cos ^2 A}{\cos ^2 A \cdot \cos ^2 B}$
$=\frac{\sin ^2 A\left(1-\sin ^2 B\right)-\sin ^2 B\left(1-\sin ^2 A\right)}{\cos ^2 A \cdot \cos ^2 B}$
$=\frac{\sin ^2 A-\sin ^2 A \cdot \sin ^2 B-\sin ^2 B+\sin ^2 A \cdot \sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}$
$=\frac{\sin ^2 A-\sin ^2 A \cdot \sin ^2 B-\sin ^2 B+\sin ^2 A \cdot \sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}$
$=\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}=\text { RHS }$
View full question & answer→
Question 1133 Marks
Prove.$\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\left(\frac{\cos A}{1+\sin A}\right)^2$
Answer
$\text { LHS }=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$
$=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1} \times \frac{\operatorname{cosec} A+1}{\operatorname{cosec} A+1}$
$=\frac{\operatorname{cosec} A-1}{(\operatorname{cosec} A+1)^2}$
$=\frac{\cot ^2 A}{(\operatorname{cosec} A+1)^2}$
$=\frac{\frac{\cos ^2 A}{\sin ^2 A}}{\left(\frac{1}{\sin A}+1\right)^2}$
$=\left(\frac{\cos A}{1+\sin A}\right)^2=\text { RHS }$
View full question & answer→
Question 1143 Marks
Prove.$(\cot A-\operatorname{cosec} A)^2=\frac{1-\cos A}{1+\cos A}$
Answer
$\text { R.H.S }=\frac{1-\cos A}{1+\cos A}$
$=\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}$
$=\frac{(1-\cos A)^2}{1-\cos ^2 A}$
$=\frac{(1-\cos A)^2}{\sin ^2 A}$
$=\left(\frac{1}{\sin A}-\frac{\cos A}{\sin A}\right)^2$
$=(\operatorname{cosec} A-\cot A)^2$
$=(\cot A-\operatorname{cosec} A)^2$
$=\text { L.H.S }$
View full question & answer→
Question 1153 Marks
Prove.$\frac{1-\sin A}{1+\sin A}=(\sec A-\tan A)^2$
Answer
$\text { LHS }=\frac{1-\sin A}{1+\sin A}$
$=\frac{1-\sin A}{1+\sin A} \times \frac{1-\sin A}{1-\sin A}$
$=\frac{(1-\sin A)^2}{1-\sin ^2 A}$
$=\frac{(1-\sin A)^2}{\cos ^2 A}$
$=\left(\frac{1-\sin A}{\cos A}\right)^2$
$=\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^2$
$=(\sec A-\tan A)^2$
$=\text { RHS }$
View full question & answer→
Question 1163 Marks
Prove.$\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2 \sec A$
Answer
$\text { LHS }=\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}$
$=\frac{(1+\sin A)^2+\cos ^2 A}{\cos A(1+\sin A)}$
$=\frac{1+\sin ^2 A+2 \sin A+\cos ^2 A}{\cos A(1+\sin A)}$
$=\frac{1+2 \sin A+1}{\cos A(1+\sin A)}$
$=\frac{2(1+\sin A)}{\cos A(1+\sin A)}$
$=2 \sec A=\text { RHS }$
View full question & answer→
Question 1173 Marks
Prove.$\frac{1+\cos A}{1-\cos A}=\frac{\tan ^2 A}{(\sec A-1)^2}$
Answer
$\begin{aligned} & \text { R.H.S }=\frac{\tan ^2 A}{(\sec A-1)^2}=\frac{\sec ^2 A-1}{(\sec A-1)^2} \quad \cdots\left[\sec ^2 \theta-\tan ^2 \theta=1\right. \\ & \left.\sec ^2 \theta-1=\tan ^2 \theta\right] \\ & =\frac{(\sec A+1)(\sec A-1)}{\sec A-1} \\ & =\frac{\sec A+1}{\sec A-1}=\frac{\frac{1}{\cos A}+1}{\frac{1}{\cos A}-1}=\frac{\frac{1+\cos A}{\cos A}}{\frac{1-\cos A}{\cos A}} \\ & =\frac{1+\cos A}{1-\cos A} \\ & \text { R.H.S. }=\text { L.H.S }\end{aligned}$
View full question & answer→
Question 1183 Marks
Prove.$\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}=2 \sec ^2 A$
Answer
$\text { LHS }=\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}$
$=\frac{\operatorname{cosec} A(\operatorname{cosec} A+1)+\operatorname{cosec} A(\operatorname{cosec} A-1)}{(\operatorname{cosec} A-1)(\operatorname{cosec} A+1)}$
$=\frac{\operatorname{cosec}{ }^2 A+\operatorname{cosec} A+\operatorname{cosec} A-\operatorname{cosec} A}{(\operatorname{cosec} A)^2-(1)^2}$
$=\frac{2 \operatorname{cosec}{ }^2 A}{\operatorname{cosec} e^2 A-1}$
$=\frac{2 \operatorname{cosec} A}{\cot ^2 A} \ldots\left(\because \operatorname{cosec}^2 A-1=\cot ^2 A \right)$
$=2 \frac{\frac{1}{\sin ^2 A}}{\cos ^2 A}$
$=\frac{\frac{\sin ^2 A}{2}}{\cos ^2 A}$
$=2 \sec ^2 A$
$=\text { RHS }$
View full question & answer→
Question 1193 Marks
prove. $(\sin\ A + cosec\ A)^2 + (\cos\ A + sec\ A)^2 = 7 + \tan^2A + \cot^2A$
Answer
$LHS =(\sin\ A + cosec\ A)^2 + (\cos\ A + sec\ A)^2$
$= \sin^2A + cosec^2A + 2 \sin\ A cosec\ A + \cos^2A + \sec^2A + 2 \cos\ A \sec\ A$
$= \sin^2A + \cos^2A +cosec^2A +\sec^2A + 2 + 2$
$= 1 +cosec^2A +\sec^2A + 4$
$= (1 + \cot^2A) + (1 + \tan^2A) + 5$
$= 7 + \tan^2A + \cot^2A = RHS$
View full question & answer→
Question 1203 Marks
Prove.$\frac{\sec A-\tan A}{\sec A+\tan A}=1-2 \sec A \tan A+2 \tan ^2 A$
Answer
$\text { LHS }=\frac{\sec A-\tan A}{\sec A+\tan A}$
$=\frac{\sec A-\tan A}{\sec A+\tan A} \times \frac{\sec A-\tan A}{\sec A-\tan A}$
$=\frac{(\sec A-\tan A)^2}{\sec ^2 A-\tan ^2 A}$
$=\frac{\sec ^2 A+\tan ^2 A-2 \sec A \tan A}{1}$
$=1+\tan ^2 A+\tan ^2 A-2 \sec A \tan A$
$=1-2 \sec A \tan A+2 \tan ^2 A=\text { RHS }$
View full question & answer→
Question 1213 Marks
Prove.$\operatorname{cosec} A+\cot A=\frac{1}{\operatorname{cosec} A-\cot A}$
Answer
$\text { LHS }=\operatorname{cosec} A+\cot A$
$=\frac{\operatorname{cosec} A+\cot A}{1} \times \frac{\operatorname{cosec} A-\cot A}{\operatorname{cosec} A-\cot A}$
$=\frac{\operatorname{cosec} c^2 A-\cot ^2 A}{\operatorname{cosec} A-\cot A}=\frac{1+\cot ^2 A-\cot ^2 A}{\operatorname{cosec} A-\cot A}$
$=\frac{1}{\operatorname{cosec} A-\cot A}=\text { RHS }$
View full question & answer→
Question 1223 Marks
Prove.
$(\operatorname{cosec} A - \sin A) (sec A - \cos A) (\tan A + \cot A) = 1$
Answer
$\text { LHS }=(\operatorname{cosec} A-\sin A )(\sec A -\cos A )(\tan A +\cot A )$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{1}{\tan A}+\tan A\right)$
$=\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right)\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)$
$=\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)\left(\frac{\sin ^2 A+\cos ^2 A}{\sin A \cdot \cos A}\right)$
$=\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)\left(\frac{1}{\sin A \cdot \cos A}\right)$
$=1=\text { RHS }$
View full question & answer→
Generate Paper FreeAll Trigonometrical Identities topics
[3 marks sum] - Page 3 - Mathematics STD 10 Questions - Vidyadip