QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
when SECA = 3 where A is an acute angle.
: tan” 60° + 4 sin® 45° + 3 sec? 60° + 5¢
Evaluaicy cos ec^30°+ sec 60° — cot? 30°
1 +tan’ A) (1 -tanA)
gE Prove that 17 + col A) (1 -cot A)*
PRIOR
MARK :
Multiple Choice Questions.
a 1
, If sin@ = then tan@ is equal to
) : b) 4s
a) —= —
45 9
If cosA = 2, then value of cotA.sinA 18:
5 HE
a) —== SURO V) :
If V^25ing = 1, then cotfxcosech is eque
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Step 1:Solution
{
"solution": {
"steps": [
{
"section": "Q^3",
"title": "Find cos A using sec A",
"formula": "\cos A = \frac{1}{\sec A} = \frac{3}{5}",
"explanation": "Since sec A is given as 5 / 3, cos A is the reciprocal."
},
{
"section": "Q^3",
"title": "Find sin A using Pythagorean identity",
"formula": "\sin^2 A = 1 - \cos^2 A",
"explanation": "Use the identity to find sin A."
},
{
"section": "Q^3",
"title": "Calculate sin A",
"formula": "\sin A = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}",
"explanation": "Substitute cos A and simplify to get sin A."
},
{
"section": "Q^3",
"title": "Final Answer",
"formula": "\sin A = \frac{4}{5}",
"explanation": "The value of sin A is 4 / 5 for the given sec A."
},
{
"section": "Q^4",
"title": "Evaluate numerator",
"formula": "\tan^2 60^\circ + 4\sin^2 45^\circ + 3\sec^2 60^\circ + 5",
"explanation": "Substitute values: tan 60° = √3, sin 45° = 1 /√2, sec 60° = 2."
},
{
"section": "Q^4",
"title": "Calculate each term in numerator",
"formula": "\tan^2 60^\circ = 3, 4\sin^2 45^\circ = 4 \times \frac{1}{2} = 2, 3\sec^2 60^\circ = 3 \times 4 = 12, 5",
"explanation": "Evaluate each trigonometric term."
},
{
"section": "Q^4",
"title": "Sum numerator terms",
"formula": "3 + 2 + 12 + 5 = 22",
"explanation": "Add all terms for the numerator."
},
{
"section": "Q^4",
"title": "Evaluate denominator",
"formula": "\csc 30^\circ + \sec 60^\circ - \cot^2 30^\circ",
"explanation": "Substitute values: csc 30° = 2, sec 60° = 2, cot 30° = √3."
},
{
"section": "Q^4",
"title": "Calculate each term in denominator",
"formula": "\csc 30^\circ = 2, \sec 60^\circ = 2, \cot^2 30^\circ = 3",
"explanation": "Evaluate each trigonometric term."
},
{
"section": "Q^4",
"title": "Sum denominator terms",
"formula": "2 + 2 - 3 = 1",
"explanation": "Add and subtract terms for the denominator."
},
{
"section": "Q^4",
"title": "Final Answer",
"formula": "\frac{22}{1} = 22",
"explanation": "The evaluated value is 22."
},
{
"section": "Q^5",
"title": "Express tan²A and cot²A in terms of tanA",
"formula": "\cot^2 A = \frac{1}{\tan^2 A}",
"explanation": "Rewrite cot²A using tanA."
},
{
"section": "Q^5",
"title": "Rewrite LHS",
"formula": "\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \tan^2 A}{1 + \frac{1}{\tan^2 A}}",
"explanation": "Substitute cot²A."
},
{
"section": "Q^5",
"title": "Simplify denominator",
"formula": "1 + \frac{1}{\tan^2 A} = \frac{\tan^2 A + 1}{\tan^2 A}",
"explanation": "Combine terms in the denominator."
},
{
"section": "Q^5",
"title": "Simplify overall expression",
"formula": "\frac{1 + \tan^2 A}{\frac{\tan^2 A + 1}{\tan^2 A}} = (1 + \tan^2 A) \times \frac{\tan^2 A}{\tan^2 A + 1} = \tan^2 A",
"explanation": "Multiply numerator by reciprocal of denominator."
},
{
"section": "Q^5",
"title": "Rewrite RHS",
"formula": "\frac{(1 - \tan A)^2}{(1 - \cot A)^2}",
"explanation": "Express RHS in terms of tanA and cotA."
},
{
"section": "Q^5",
"title": "Express cotA in terms of tanA",
"formula": "\cot A = \frac{1}{\tan A}",
"explanation": "Substitute cotA."
},
{
"section": "Q^5",
"title": "Rewrite denominator of RHS",
"formula": "(1 - \cot A)^2 = \left(1 - \frac{1}{\tan A}\right)^2 = \left(\frac{\tan A - 1}{\tan A}\right)^2",
"explanation": "Express denominator in terms of tanA."
},
{
"section": "Q^5",
"title": "Rewrite numerator of RHS",
"formula": "(1 - \tan A)^2",
"explanation": "Numerator remains as is."
},
{
"section": "Q^5",
"title": "Combine numerator and denominator",
"formula": "\frac{(1 - \tan A)^2}{\left(\frac{\tan A - 1}{\tan A}\right)^2} = (1 - \tan A)^2 \times \frac{\tan^2 A}{(\tan A - 1)^2}",
"explanation": "Multiply numerator by reciprocal of denominator."
},
{
"section": "Q^5",
"title": "Simplify RHS",
"formula": "\tan^2 A",
"explanation": "The terms (1 - tanA)^2 and (tanA - 1)^2 cancel out, leaving tan²A."
},
{
"section": "Q^5",
"title": "Final Answer",
"formula": "\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{(1 - \tan A)^2}{(1 - \cot A)^2}",
"explanation": "Both sides simplify to tan²A, hence proved."
},
{
"section": "MCQ^1",
"title": "Given sin θ = 1 / 9, find tan θ",
"formula": "\sin \theta = \frac{1}{9}",
"explanation": "Let the opposite side be 1, hypotenuse be 9."
},
{
"section": "MCQ^1",
"title": "Find adjacent side using Pythagoras",
"formula": "\text{adjacent} = \sqrt{9^2 - 1^2} = \sqrt{81 - 1} = \sqrt{80} = 4\sqrt{5}",
"explanation": "Calculate the adjacent side."
},
{
"section": "MCQ^1",
"title": "Calculate tan θ",
"formula": "\tan \theta = \frac{1}{4\sqrt{5}}",
"explanation": "Tan θ is opposite over adjacent."
},
{
"section": "MCQ^1",
"title": "Final Answer",
"formula": "\tan \theta = \frac{1}{4\sqrt{5}}",
"explanation": "Correct option is (a)."
},
{
"section": "MCQ^2",
"title": "Given cos A = 5 / 8, find sin A",
"formula": "\sin^2 A = 1 - \cos^2 A",
"explanation": "Use Pythagorean identity."
},
{
"section": "MCQ^2",
"title": "Calculate sin A",
"formula": "\sin A = \sqrt{1 - \left(\frac{5}{8}\right)^2} = \sqrt{1 - \frac{25}{64}} = \sqrt{\frac{39}{64}} = \frac{\sqrt{39}}{8}",
"explanation": "Find sin A."
},
{
"section": "MCQ^2",
"title": "Find cot A",
"formula": "\cot A = \frac{\cos A}{\sin A} = \frac{5 / 8}{\sqrt{39}/ 8} = \frac{5}{\sqrt{39}}",
"explanation": "Cot A is cos A over sin A."
},
{
"section": "MCQ^2",
"title": "Calculate cotA.sinA",
"formula": "\cot A \cdot \sin A = \frac{5}{\sqrt{39}} \cdot \frac{\sqrt{39}}{8} = \frac{5}{8}",
"explanation": "Multiply cot A and sin A."
},
{
"section": "MCQ^2",
"title": "Final Answer",
"formula": "\cot A \cdot \sin A = \frac{5}{8}",
"explanation": "Correct option is (b)."
},
{
"section": "MCQ^3",
"title": "Given $\sqrt{2} \sin \theta = 1$, find sin θ",
"formula": "\sin \theta = \frac{1}{\sqrt{2}}",
"explanation": "Divide both sides by √2."
},
{
"section": "MCQ^3",
"title": "Find cosec θ",
"formula": "\csc \theta = \frac{1}{\sin \theta} = \sqrt{2}",
"explanation": "Cosec θ is reciprocal of sin θ."
},
{
"section": "MCQ^3",
"title": "Find cot θ",
"formula": "\cot \theta = \frac{\cos \theta}{\sin \theta}",
"explanation": "Use definition of cot θ."
},
{
"section": "MCQ^3",
"title": "Find cos θ using Pythagoras",
"formula": "\cos^2 \theta = 1 - \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2}, \cos \theta = \frac{1}{\sqrt{2}}",
"explanation": "Calculate cos θ."
},
{
"section": "MCQ^3",
"title": "Calculate cot θ",
"formula": "\cot \theta = \frac{1 /\sqrt{2}}{1 /\sqrt{2}} = 1",
"explanation": "Cot θ is cos θ over sin θ."
},
{
"section": "MCQ^3",
"title": "Calculate cot θ × cosec θ",
"formula": "1 \times \sqrt{2} = \sqrt{2}",
"explanation": "Multiply cot θ and cosec θ."
},
{
"section": "MCQ^3",
"title": "Final Answer",
"formula": "\sqrt{2}",
"explanation": "The value is √2."
}
]
}
}
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