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Maths 2 Marks Questions

Trigonometric Ratios of Complementary Angles · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 13 questions.

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Question 12 Marks
Prove that:
$\text{cosec}(65^\circ+\theta)-\sec(25^\circ-\theta)\\-\tan(55^\circ-\theta)+\cot(35^\circ+\theta)=0$
Answer
$\text{L.H.S.}=\text{cosec}(65^\circ+\theta)-\sec(25^\circ-\theta)\\ \ \ -\tan(55^\circ-\theta)+\cot(35^\circ+\theta)$
$=\text{cosec}\big\{90^\circ-(25^\circ-\theta)\big\}-\sec(25^\circ-\theta)\\ \ \ -\tan(55^\circ-\theta)+\cot\big\{90^\circ-(55^\circ-\theta)\big\}$
$=\sec(25^\circ+\theta)-\sec(25^\circ-\theta)\\-\tan(55^\circ-\theta)+\tan(55^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
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Question 22 Marks
Prove that:
$\frac{\sin\theta\cos(90^\circ-\theta)\cos\theta}{\sin(90^\circ-\theta)}+\frac{\cos\theta\sin(90^\circ-\theta)\sin\theta}{\cos(90^\circ-\theta)}=1$
Answer
$\text{L.H.S.}=\frac{\sin\theta\cos(90^\circ-\theta)\cos\theta}{\sin(90^\circ-\theta)}+\frac{\cos\theta\sin(90^\circ-\theta)\sin\theta}{\cos(90^\circ-\theta)}$
$=\frac{\sin\theta\sin\theta\cos\theta}{\cos\theta}+\frac{\cos\theta\cos\theta\sin\theta}{\sin\theta}$
$=\sin^2\theta+\cos^2\theta$
$=1$
$=\text{R.H.S.}$
Hence Proved.
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Question 32 Marks
Prove that:
$\sin\theta\cos(90^\circ-\theta)+\sin(90^\circ-\theta)\cos\theta=1$
Answer
$\text{L.H.S.}=\sin\theta\cos(90^\circ-\theta)+\sin(90^\circ-\theta)\cos\theta$
$=\sin\theta\sin\theta+\cos\theta\cos\theta$
$=\sin^2\theta+\cos^2\theta$
$=1$
$=\text{R.H.S.}$
Hence Proved.
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Question 42 Marks
Prove that:
$\frac{\sin18^\circ}{\cos72^\circ}+\sqrt{3}\big(\tan10^\circ\tan30^\circ\tan40^\circ\tan50^\circ\tan80^\circ\big)=2$
Answer
$\text{L.H.S.}=\frac{\sin18^\circ}{\cos72^\circ} +\sqrt{3}\big(\tan10^\circ\tan30^\circ\tan40^\circ\tan50^\circ\tan80^\circ\big)$
$=\frac{\sin18^\circ}{\sin(90^\circ-72^\circ)} \\+\sqrt{3}\Big[\cot(90^\circ-10^\circ)\times\frac{1}{\sqrt{3}}\times\cot(90^\circ-40^\circ)\times\tan50^\circ\times\tan80^\circ\Big]$
$=\frac{\sin18^\circ}{\sin18^\circ}+\sqrt{3}\Big(\frac{\cot80^\circ\times\cot50^\circ\times\tan80^\circ}{\sqrt{3}}\Big)$
$=1+\Big(\frac{1}{\tan80^\circ}\times\frac{1}{\tan50^\circ}\times\tan50^\circ\times\tan80^\circ\Big)$
$=1+1$
$=2$
$=\text{R.H.S.}$
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Question 52 Marks
Prove that:
$\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec }31^\circ=2$
Answer
$\text{L.H.S}=\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec }31^\circ$
$=\frac{\cos80^\circ}{\cos(90^\circ-10^\circ)}+\sin(90^\circ-59^\circ)\text{cosec }31^\circ$
$=\frac{\cos80^\circ}{\cos80^\circ}+\sin31^\circ\text{cosec }31^\circ$
$=1+\sin31^\circ\times\frac{1}{\sin31^\circ}$
$=1+1$
$=2$
$=\text{R.H.S.}$
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Question 62 Marks
Prove that:
$\frac{\sec(90^\circ-\theta)\text{cosec }\theta-\tan(90^\circ-\theta)\cot\theta+\cos^225^\circ+\cos^265^\circ}{3\tan27^\circ\tan63^\circ}=\frac{2}{3}$
Answer
$\text{L.H.S.}=\frac{\sec(90^\circ-\theta)\text{cosec }\theta-\tan(90^\circ-\theta)\cot\theta+\cos^225^\circ+\cos^265^\circ}{3\tan27^\circ\tan63^\circ}$
$=\frac{\text{cosec }\theta\text{ cosec }\theta-\cot\theta\cot\theta+\sin^2(90^\circ-25^\circ)+\cos^265^\circ}{3\tan27^\circ\cot27^\circ}$
$=\frac{1+1}{3\times\tan27^\circ\times\frac{1}{\tan27^\circ}}$
$=\frac23$
$=\text{R.H.S.}$
Hence Proved.
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Question 72 Marks
Prove that:
$\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec 20}^\circ}{\sec70^\circ}-2\cos70^\circ\text{cosec }20^\circ=0$
Answer
$\text{L.H.S}=\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec 20}^\circ}{\sec70^\circ}-2\cos70^\circ\text{cosec }20^\circ$
$=\frac{\sin70^\circ}{\sin(90^\circ-20^\circ)}+\frac{\sec(90^\circ-20^\circ)}{\sec70^\circ}-2\cos70^\circ\sec70^\circ$
$=\frac{\sin70^\circ}{\sin70^\circ}+\frac{\sec70^\circ}{\sec70^\circ}-2\cos70^\circ\sec70^\circ$
$=1+1-2\times\cos70^\circ\times\frac{1}{\cos70^\circ}$
$=2-2$
$=0$
$=\text{R.H.S.}$
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Question 82 Marks
Prove that:
$\text{cosec}(67^\circ+\theta)-\sec(23^\circ-\theta)=0$
Answer
$\text{L.H.S.}=\text{cosec}(67^\circ+\theta)-\sec(23^\circ-\theta)$
$=\text{cosec}\big\{90^\circ-(23^\circ-\theta)\big\}-\sec(23^\circ-\theta)$
$=\sec(23^\circ+\theta)-\sec(23^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
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Question 92 Marks
Prove that:
$=\frac{2\sin68^\circ}{\cos22^\circ}-\frac{2\cot15^\circ}{5\tan75^\circ}\\-\frac{3\tan45^\circ\tan20^\circ\tan40^\circ\tan50^\circ\tan70^\circ}{5}=1$
Answer
$\text{L.H.S}=\frac{2\sin68^\circ}{\cos22^\circ}-\frac{2\cot15^\circ}{5\tan75^\circ}\\ \ -\frac{3\tan45^\circ\tan20^\circ\tan40^\circ\tan50^\circ\tan70^\circ}{5}$
$=\frac{2\sin68^\circ}{\sin(90^\circ-22^\circ)}-\frac{2\cot15^\circ}{5\cot(90^\circ-75^\circ)}\\ \ \ -\frac{3\times1\times\cot(90^\circ-20^\circ)\times\cot(90^\circ-40^\circ)\times\tan50^\circ\times\tan70^\circ}{5}$
$=\frac{2\sin68^\circ}{\sin68^\circ}-\frac{2\cot15^\circ}{5\cot15^\circ}\\ \ \ -\frac{3\cot70^\circ\cot50^\circ\tan50^\circ\tan70^\circ}{5}$
$=2-\frac25-\frac{3\times\frac{1}{\tan70^\circ}\times\frac{1}{\tan50^\circ}\times\tan50^\circ\times\tan70^\circ}{5}$
$=2-\frac{2}{5}-\frac35$
$=\frac{10-2-3}{5}$
$=\frac55$
$=1$
$=\text{R.H.S.}$
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Question 102 Marks
Prove that:
$\tan(55^\circ-\theta)-\cot(35^\circ+\theta)=0$
Answer
$\text{L.H.S.}=\tan(55^\circ-\theta)-\cot(35^\circ+\theta)$
$=\tan\big\{90^\circ-(35^\circ-\theta)\big\}-\cot(35^\circ-\theta)$
$=\cot(35^\circ+\theta)-\cot(35^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
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Question 112 Marks
Prove that:
$\frac{7\cos55^\circ}{3\sin35^\circ}-\frac{4(\cos70^\circ\text{cosec }20^\circ)}{3(\tan5^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan85^\circ)}=1$
Answer
$\text{L.H.S.}=\frac{7\cos55^\circ}{3\sin35^\circ}\\ \ \ -\frac{4(\cos70^\circ\text{cosec }20^\circ)}{3(\tan5^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan85^\circ)}$
$=\frac{7\cos55^\circ}{3\cos(90^\circ-35^\circ)}\\ \ -\frac{4\big[\sin(90^\circ-70^\circ)\text{cosec }20^\circ\big]}{3\big[\cot(90^\circ-5^\circ)\times\cot(90^\circ-25^\circ)\times1\times\tan65^\circ\times\tan85^\circ\big]}$
$=\frac{7\cos55^\circ}{3\cos55^\circ}-\frac{4(\sin(20^\circ\text{cosec }20^\circ)}{3(\cot85^\circ\cot65^\circ\tan65^\circ\tan85^\circ)}$
$=\frac73-\frac{4\Big(\sin20^\circ\times\frac{1}{\sin20^\circ}\Big)}{3\Big(\frac{1}{\tan85^\circ}\times\frac{1}{\tan65^\circ}\times\tan65^\circ\times\tan85^\circ\Big)}$
$=\frac73-\frac43$
$=\frac33$
$=1$
$=\text{R.H.S.}$
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Question 122 Marks
Prove that:
$\sin(50^\circ+\theta)-\cos(40^\circ-\theta)\\+\tan1^\circ+\tan10^\circ\tan80^\circ\tan89^\circ=1$
Answer
$\text{L.H.S.}=\sin(50^\circ+\theta)-\cos(40^\circ-\theta)\\ \ \ +\tan1^\circ+\tan10^\circ\tan80^\circ\tan89^\circ$
$=\sin\big\{90^\circ-(40^\circ-\theta)\big\}-\cos(40^\circ-\theta)\\ \ \ +\big\{\tan1^\circ\tan(90^\circ-1^\circ)\big\}\big\{\tan10^\circ\tan(90^\circ-10)\big\}$
$=\cos(40^\circ-\theta)-\cos(40^\circ-\theta)\\+(\tan1^\circ\cot1^\circ)(\tan10^\circ\cot10^\circ)$
$=\Big(\frac{1}{\cot1^\circ}\times\cot1^\circ\Big)\Big(\tan10^\circ\times\frac{1}{\tan10^\circ}\Big)$
$=1\times1$
$=1$
$=\text{R.H.S.}$
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Question 132 Marks
Prove that:
$\frac{\sin\theta}{\cos(90^\circ-\theta)}+\frac{\cos\theta}{\sin(90^\circ-\theta)}=2$
Answer
$\text{L.H.S.}=\frac{\sin\theta}{\cos(90^\circ-\theta)}+\frac{\cos\theta}{\sin(90^\circ-\theta)}$
$=\frac{\sin\theta}{\sin\theta}+\frac{\cos\theta}{\cos\theta}$
$=1+1$
$=2$
$=\text{R.H.S.}$
Hence Proved.
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