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Mathematics [5 marks sum]

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

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[5 marks sum]

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Question 15 Marks
If A and B are complementary angles, prove that:$\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\cos B-\cos A}{\cos B+\cos A}=\frac{2}{2 \sin ^2 A-1}$
Answer
Since, $A$ and $B$ are complementary angles, $A+B=90^{\circ}$
$\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\cos B-\cos A}{\cos B+\cos A}$
$ =\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\cos \left(90^{\circ}-A\right)-\cos \left(90^{\circ}-B\right)}{\cos \left(90^{\circ}-A\right)+\cos \left(90^{\circ}-B\right)} $
$ =\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\sin A-\sin B}{\sin A+\sin B} $
$ =\frac{(\sin A+\sin B)^2+(\sin A-\sin B)^2}{(\sin A-\sin B)(\sin A+\sin B)} $
$=\frac{\sin ^2 A+\sin ^2 B+2 \sin A \sin B+\sin ^2 A+\sin ^2 B-2 \sin A}{\sin ^2 A-\sin ^2 B} $
$ =2 \frac{\sin ^2 A+\sin ^2 B}{\sin ^2 A-\sin ^2 B} $
$ =2 \frac{\sin ^2 A+\sin ^2\left(90^{\circ}-A\right)}{\sin ^2 A-\sin ^2\left(90^{\circ}-A\right)} $
$=2 \frac{\sin ^2 A+\cos ^2 B}{\sin ^2 A-\cos ^2 B} $
$ =\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)} $
$ =\frac{2}{2 \sin ^2 A-1} $
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Question 25 Marks
If $\tan A = n \tan B$ and $\sin A = m \sin B,$ prove that:
$\cos ^2 A=\frac{m^2-1}{n^2-1}$
Answer
Given that, $\tan A = n \tan B$ and $\sin A = m \sin B.$
$\Rightarrow n=\frac{\tan A}{\tan B}$ and $m=\frac{\sin A}{\sin B}$
$\therefore \frac{m^2-1}{n^2-1} $
$ =\frac{\left(\frac{\sin A}{\sin B}\right)^2-1}{\left(\frac{\tan A}{\tan B}\right)^2-1} $
$=\frac{\tan ^2 B\left(\sin ^2 A-\sin ^2 B\right)}{\sin ^2 B\left(\tan ^2 A-\tan ^2 B\right)}$
$ =\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 B\left(\frac{\sin ^2 A}{\cos ^2 A}-\frac{\sin ^2 B}{\cos ^2 B}\right)} $
$=\frac{\cos ^2 A\left(\sin ^2 A-\sin ^2 B\right)}{\sin ^2 A \cos ^2 B-\left(1-\cos ^2 B\right) \cos ^2 A}$
$ =\frac{\cos ^2 A\left(1-\cos ^2 A-1+\cos ^2 B\right)}{\cos ^2 B\left(\sin ^2 A+\cos ^2 A\right)-\cos ^2 A} $
$=\frac{\cos ^2 A\left(\cos ^2 B-\cos ^2 A\right)}{\cos ^2 B-\cos ^2 A} $
$ =\cos ^2 A$
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Question 35 Marks
Prove the following identitie:
$\cot ^2 A\left(\frac{\sec A-1}{1+\sin A}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)=0$
Answer
$\cot ^2 A\left(\frac{\sec A-1}{1+\sin A}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right) $
$ =\cot ^2 A\left(\frac{\sec A-1}{1+\sin A} \times \frac{\sec A+1}{\sec A+1}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right) $
$ =\cot ^2 A\left(\frac{\sec ^2 A-1}{(1+\sin A)(\sec A+1)}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right) $
$ =\cot ^2 A\left(\frac{\tan ^2 A}{(1+\sin A)(\sec A+1)}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right) $
$=\frac{1}{(1+\sin A)(\sec A+1)}+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right) $
$ =\frac{1+\sec ^2 A(\sin A-1)\left(1+\sin ^2 A\right)}{(1+\sin A)(\sec A+1)} $
$=\frac{1+\sec ^2 A(\sin 2-1)}{(1+\sin A)(\sec A+1)} $
$ =\frac{1+\sec ^2 A\left(-\cos ^2 A\right)}{(1+\sin A)(\sec A+1)}$
$=\frac{1-1}{(1+\sin A)(\sec A+1)} $
$ =0$
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Question 45 Marks
Prove the following identitie:$\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{\cos A}{1-\sin A}$
Answer
$\frac{\sin A-\cos A+1}{\sin A+\cos A-1} $
$ =\frac{\sin A-\cos A+1}{\sin A+\cos A-1} \times \frac{\sin A-(\cos A-1)}{\sin A-(\cos A-1)} $
$ =\frac{(\sin A-\cos A+1)^2}{\sin ^2 A-(\cos A-1)^2}$
$ =\frac{\sin ^2 A+\cos ^2 A+1-2 \sin A \cos A-2 \cos A+2 \sin A}{\sin ^2 A-\cos ^2 A-1+2 \cos A}$
$ =\frac{1+1-2 \sin A \cos A-2 \cos A+2 \sin A}{-\cos ^2 A-\cos ^2 A+2 \cos A} $
$=\frac{2(1-\cos A)+2 \sin A(1-\cos A)}{2 \cos ^2(1-\cos A)} $
$ =\frac{1+\sin A}{\cos A} $
$ =\frac{1+\sin A}{\cos A} \times \frac{1-\sin A}{1-\sin A} $
$ =\frac{\cos ^2 A}{\cos A(1-\sin A)} $
$=\frac{\cos A}{1-\sin A}$
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Question 55 Marks
Prove that
$\cot ^2 A-\cot ^2 B=\frac{\cos ^2 A-\cos ^2 B}{\sin ^2 A \sin ^2 B}=\operatorname{cosec}{ }^2 A-\operatorname{cosec} B$
Answer
$\cot ^2 A-\cot ^2 B $
$ =\frac{\cos ^2 A}{\sin ^2 A}-\frac{\cos ^2 B}{\sin ^2 B} $
$ =\frac{\cos ^2 A \sin ^2 B-\cos ^2 B \sin ^2 A}{\sin ^2 A \sin ^2 B} $
$ =\frac{\cos ^2 A\left(1-\cos ^2 B\right)-\cos ^2 B\left(1-\cos ^2 A\right)}{\sin ^2 A \sin ^2 B} $
$ =\frac{\cos ^2 A-\cos ^2 A \cos ^2 B-\cos ^2 B+\cos ^2 B \cos ^2 A}{\sin ^2 A \sin ^2 B} $
$ =\frac{\cos ^2 A-\cos ^2 B}{\sin ^2 A \sin ^2 B} $
$=\frac{1-\sin ^2 A-1+\sin ^2 B}{\sin ^2 A \sin ^2 B}$
$=\frac{-\sin ^2 A+\sin ^2 B}{\sin ^2 A \sin ^2 B} $
$=\frac{\sin ^2 B}{\sin ^2 A \sin B}-\frac{\sin ^2 A}{\sin ^2 A \sin ^2 B}$
$=\frac{1}{\sin ^2 A}-\frac{1}{\sin ^2 B} $
$ =\operatorname{cosec}^2 A-\operatorname{cosec}^2 B$
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Question 65 Marks
If $A$ and $B$ are complementary angles, prove that:
$\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\cos B-\cos A}{\cos B+\cos A}=\frac{2}{2 \sin ^2 A-1}$
Answer
Since, $A$ and $B$ are complementary angles, $A+B=90^{\circ}$
$\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\cos B-\cos A}{\cos B+\cos A}$
$=\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\cos \left(90^{\circ}-A\right)-\cos \left(90^{\circ}-B\right)}{\cos \left(90^{\circ}-A\right)+\cos \left(90^{\circ}-B\right)}$
$=\frac{\sin A+\sin B}{\sin A-\sin B}+\frac{\sin A-\sin B}{\sin A+\sin B}$
$=\frac{(\sin A+\sin B)^2+(\sin A-\sin B)^2}{(\sin A-\sin B)(\sin A+\sin B)}$
$=\frac{\sin ^2 A+\sin ^2 B+2 \sin A \sin B+\sin ^2 A+\sin ^2 B-2 \sin A}{\sin ^2 A-\sin ^2 B}$
$=2 \frac{\sin ^2 A+\sin ^2 B}{\sin ^2 A-\sin ^2 B}$
$=2 \frac{\sin ^2 A+\sin ^2\left(90^{\circ}-A\right)}{\sin ^2 A-\sin ^2\left(90^{\circ}-A\right)}$
$=2 \frac{\sin ^2 A+\cos ^2 B}{\sin ^2 A-\cos ^2 B}$
$=\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)}$
$=\frac{2}{2 \sin ^2 A-1}$
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Question 75 Marks
If $\tan A=n \tan B$ and $\sin A=m \sin B$, prove that:
$\cos ^2 A=\frac{m^2-1}{n^2-1}$
Answer
$\begin{aligned} & \text { Given that, } \tan A=n \tan B \text { and } \sin A=m \sin B \text {. } \\ & \Rightarrow n=\frac{\tan A}{\tan B} \text { and } m=\frac{\sin A}{\sin B} \\ & \therefore \frac{m^2-1}{n^2-1}=\frac{\left(\frac{\sin A}{\sin B}\right)^2-1}{\left(\frac{\tan A}{\tan B}\right)^2-1} \\ & =\frac{\frac{\sin ^2 A}{\sin ^2 B}-\frac{1}{1}}{\frac{\tan ^2 A}{\tan ^2 B}-1} \\ & =\frac{\frac{\sin ^2 A-\sin ^2 B}{\sin ^2 B}}{\frac{\tan ^2 A-\tan ^2 B}{\tan ^2 B}} \\ & =\frac{\frac{\sin ^2 A-\sin ^2 B}{\tan ^2 B}}{\frac{\tan ^2 A-\tan ^2 B}{\sin ^2 B}} \\ & =\frac{\left(\sin ^2 A-\sin ^2 B\right) \sin ^2 B}{\left(\frac{\sin ^2 A}{\cos ^2 A}-\frac{\sin ^2 B}{\cos ^2 B}\right) \cos ^2 B \sin ^2 B} \\ & =\frac{\sin ^2 A-\sin ^2 B}{\left(\frac{\sin ^2 A \cdot \cos ^2 B-\sin ^2 B \cdot \cos ^2 A}{\cos ^2 A \cdot \cos ^2 B}\right) \cos ^2 B} \\ & =\frac{\left(\sin ^2 A-\sin ^2 B\right) \cos ^2 A}{\sin ^2 A \cdot \cos ^2 B-\sin ^2 B \cdot \cos ^2 A} \\ & =\frac{\left(\sin ^2 A-\sin ^2 B\right) \cos ^2 A}{\sin ^2 A\left(1-\sin ^2 B\right)-\sin ^2 B\left(1-\sin ^2 A\right)} \\ & =\frac{\left(\sin ^2 A-\sin ^2 B\right) \cos ^2 A}{\sin ^2 A-\sin ^2 A \cdot \sin ^2 B-\sin ^2 B+\sin ^2 B \cdot \sin ^2 A} \\ & =\frac{\left(\sin ^2 A-\sin ^2 B\right) \cos ^2 A}{\sin ^2 A-\sin ^2 B} \\ & =\cos ^2 A \\ & \end{aligned}$
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Question 85 Marks
Prove the following identitie:$\cot ^2 A\left(\frac{\sec A-1}{1+\sin A}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)=0$
Answer
$\cot ^2 A\left(\frac{\sec A-1}{1+\sin A}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)$
$=\cot ^2 A\left(\frac{\sec A-1}{1+\sin A} \times \frac{\sec A+1}{\sec A+1}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)$
$=\cot ^2 A\left(\frac{\sec ^2 A-1}{(1+\sin A)(\sec A+1)}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)$
$=\cot ^2 A\left(\frac{\tan ^2 A}{(1+\sin A)(\sec A+1)}\right)+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)$
$=\frac{1}{(1+\sin A)(\sec A+1)}+\sec ^2 A\left(\frac{\sin A-1}{1+\sec A}\right)$
$=\frac{1+\sec ^2 A(\sin A-1)\left(1+\sin ^2 A\right)}{(1+\sin A)(\sec A+1)}$
$=\frac{1+\sec ^2 A(\sin 2-1)}{(1+\sin A)(\sec A+1)}$
$=\frac{1+\sec ^2 A\left(-\cos ^2 A\right)}{(1+\sin A)(\sec A+1)}$
$=\frac{1-1}{(1+\sin A)(\sec A+1)}$
$=0$
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Question 95 Marks
Prove the following identitie:
$\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{\cos A}{1-\sin A}$
Answer
$\frac{\sin A-\cos A+1}{\sin A+\cos A-1}$
$=\frac{\sin A-\cos A+1}{\sin A+\cos A-1} \times \frac{\sin A-(\cos A-1)}{\sin A-(\cos A-1)}$
$=\frac{(\sin A-\cos A+1)^2}{\sin ^2 A-(\cos A-1)^2}$
$=\frac{\sin ^2 A+\cos ^2 A+1-2 \sin A \cos A-2 \cos A+2 \sin A}{\sin ^2 A-\cos ^2 A-1+2 \cos A}$
$=\frac{1+1-2 \sin A \cos A-2 \cos A+2 \sin A}{-\cos ^2 A-\cos ^2 A+2 \cos A}$
$=\frac{2(1-\cos A)+2 \sin A(1-\cos A)}{2 \cos ^2(1-\cos A)}$
$=\frac{1+\sin A}{\cos A}$
$=\frac{1+\sin A}{\cos A} \times \frac{1-\sin A}{1-\sin A}$
$=\frac{\cos ^2 A}{\cos A(1-\sin A)}$
$=\frac{\cos A}{1-\sin A}$
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Question 105 Marks
Prove that$\cot ^2 A-\cot ^2 B=\frac{\cos ^2 A-\cos ^2 B}{\sin ^2 A \sin ^2 B}=\operatorname{cosec}^2 A-\operatorname{cosec} e^2 B$
Answer
$\cot ^2 A-\cot ^2 B$
$=\frac{\cos ^2 A}{\sin ^2 A}-\frac{\cos ^2 B}{\sin ^2 B}$
$=\frac{\cos ^2 A \sin ^2 B-\cos ^2 B \sin ^2 A}{\sin ^2 A \sin ^2 B}$
$=\frac{\cos ^2 A\left(1-\cos ^2 B\right)-\cos ^2 B\left(1-\cos ^2 A\right)}{\sin ^2 A \sin ^2 B}$
$=\frac{\cos ^2 A-\cos ^2 A \cos ^2 B-\cos ^2 B+\cos ^2 B \cos ^2 A}{\sin ^2 A \sin ^2 B}$
$=\frac{\cos ^2 A-\cos ^2 B}{\sin ^2 A \sin ^2 B}$
$=\frac{1-\sin ^2 A-1+\sin ^2 B}{\sin ^2 A \sin ^2 B}$
$=\frac{-\sin ^2 A+\sin ^2 B}{\sin ^2 A \sin ^2 B}$
$=\frac{\sin ^2 B}{\sin ^2 A \sin B}-\frac{\sin ^2 A}{\sin ^2 A \sin ^2 B}$
$=\frac{1}{\sin ^2 A}-\frac{1}{\sin ^2 B}$
$=\operatorname{cosec}^2 A-\operatorname{cosec}^2 B$
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[5 marks sum] - Mathematics STD 10 Questions - Vidyadip