![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\nQuestion 1.<\/p>\n
(A) -1
\n(B) 0
\n(C) 1
\n(D) \\(\\frac{3}{2}\\)
\nAnswer:
\n(B) 0<\/p>\n
Explanation:
\ncosec (75\u00b0 + 0) – sec (15\u00b0 – 0) – tan (55\u00b0 + 0) + cot (35\u00b0-0)
\n= cosec [90\u00b0 – (15\u00b0 – 0)] – sec(15\u00b0 – 0)
\n– tan(55\u00b0 + 0) + cot[90\u00b0 – (55\u00b0 + 0)]
\n= sec(15\u00b0 – 0) – sec(15\u00b0 – 0) – tan(55\u00b0 + 0)
\n+ tan(55\u00b0 + 0)
\n= 0<\/p>\n
<\/p>\n
Question 2.<\/p>\n
(A) cos \u00df
\n(B) cos 2\u00df
\n(C) sin \u03b1
\n(D) sin 2\u03b1
\nAnswer:
\n(B) cos 2\u00df<\/p>\n
Explanation:
\ncos (\u03b1 + \u00df) = 0
\ncos(\u03b1 + \u00df) = cos 90\u00b0
\n\u03b1 + \u00df = 90\u00b0
\na = 90\u00b0 – \u00df
\nsin (\u03b1 – \u00df) = sin (90\u00b0 – \u00df – \u00df
\n= sin (90\u00b0 – 2\u00df)
\n= cos 2\u00df<\/p>\n
Question 3.<\/p>\n
(A) 0
\n(B) 1
\n(C) 2
\n(D) \\(\\frac{1}{2}\\)
\nAnswer:
\n(B) 1<\/p>\n
Explanation:
\n(tan 1\u00b0 tan 2\u00b0 tan 3\u00b0 … tan 89\u00b0)
\n= (tan 1\u00b0 tan 89\u00b0)(tan 2\u00b0 tan 88\u00b0)(tan 3\u00b0 tan 87\u00b0)…(tan 45\u00b0 tan 45\u00b0)
\n= [tan 1\u00b0 tan (90\u00b0-l)][tan 2\u00b0 tan (90\u00b0 -2)][tan 3\u00b0tan (90\u00b0-3)] … [tan 45\u00b0 tan (90\u00b0-45\u00b0)] = tan l\u00b0cot 1\u00b0 tan 2\u00b0cot 2\u00b0 tan 3\u00b0cot 3\u00b0…tan 45\u00b0cot 45\u00b0
\n= tan 1\u00b0 \u00d7 \\(\\frac{1}{\\tan 1^{\\circ}}\\)tan 2\u00b0. \\(\\frac{1}{\\tan 2^{\\circ}}\\)tan 3\u00b0. \\(\\frac{1}{\\tan 3^{\\circ}}\\) … \\(\\frac{\\tan 45^{\\circ}}{\\tan 45^{\\circ}}\\)
\n= 1 . 1 . 1 . 1…… 1 . 1
\n= 1<\/p>\n
Question 4.<\/p>\n
(A) \\(\\frac{1}{\\sqrt{3}}\\)
\n(B) \u221a 3
\n(C) 1
\n(D) 0
\nAnswer:
\n(C) 1<\/p>\n
Explanation:
\ncos 9\u03b1 = sin \u03b1
\ncos 9\u03b1 = cos (90\u00b0 – \u03b1 )
\nOn comparing both sides, we have
\n9\u03b1 = 90\u00b0 – \u03b1
\n10 \u03b1 = 90\u00b0
\n\u03b1 = 9\u00b0
\n\u2234 tan 5 \u00d7 9\u00b0 = tan 45\u00b0 = 1<\/p>\n
Question 5.<\/p>\n
(A) 0
\n(B) 1
\n(C) \\(\\frac{1}{2}\\)
\n(D) \\(\\frac{\\sqrt{3}}{2}\\)
\nAnswer:
\n(A) 0<\/p>\n
Explanation:
\nWe know that, in \u2206ABC
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Introduction-to-Trigonometry-1.png?resize=186%2C117&ssl=1\")
\nsum of three angles = 180\u00b0
\ni.e., \u2220 A + \u2220B + \u2220C = 180\u00b0
\n\u2220C = 90\u00b0
\n\u2220A +\u2220 B + 90\u00b0 = 180\u00b0
\nA + B = 90\u00b0
\n\u2234 cos(A + B) = cos 90\u00b0 = 0<\/p>\n
Question 6.<\/p>\n
(A) 0\u00b0
\n(B) 30\u00b0
\n(C) 60\u00b0
\n(D) 90\u00b0
\nAnswer:
\n(D) 90\u00b0
\nGiven, sin \u03b8 = \\(\\frac{1}{2}\\) = sin 30\u00b0
\n[ \u2234 sin 30\u00b0 = \\(\\frac{1}{2}\\)]
\n\u2234 \u03b8 = 30\u00b0
\nAnd, cos\u00df = \\(\\frac{1}{2}\\) = cos 60\u00b0
\n\u2234 \u00df = 60\u00b0
\n\u2234 \u03b8 + \u00df = 30\u00b0 + 60\u00b0 = 90\u00b0
\n[ \u2234 cos 60\u00b0 = \\(\\frac{1}{2}\\)<\/p>\n
Question 7.<\/p>\n
(A) 3
\n(B) 2
\n(C) 1
\n(D) 0
\nAnswer:
\n(B) 2<\/p>\n
Explanation: Question 8.<\/p>\n (A) \\(\\frac{2}{3}\\) Explanation: Question 9.<\/p>\n (A) \\(\\frac{3}{5}\\) Explanation: Question 10.<\/p>\n (A) \u221a3 Explanation: Question 11.<\/p>\n (A) \\(\\frac{b}{\\sqrt{b^{2}-a^{2}}}\\) Explanation: Question 12.<\/p>\n (A) 1 Explanation: Question 13.<\/p>\n (A) 1 Explanation: Question 14<\/p>\n (A) 0 Explanation: Question 15.<\/p>\n (A) sec A Explanation: Assertion and Reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 3.<\/p>\n Answer: Explanation: Question 4.<\/p>\n Answer: Explanation: Question 5.<\/p>\n Answer: Explanation: Question 6.<\/p>\n Answer: Expianation : Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) 45\u00b0 Expianation : Question 2.<\/p>\n (A) \u221a3 Explanation: Question 3.<\/p>\n (A) 400 m Explanation: Question 4.<\/p>\n (A) 1 Explanation: Question 5.<\/p>\n (A) 1 Explanation: II. Read the following text and answer the following question on the basis of the same: Question 1.<\/p>\n (A) 8 m Explanation: Question 2.<\/p>\n (A) \\(\\frac{1}{4}\\) Question 3<\/p>\n (A) 5 Explanation: Question 4<\/p>\n (A) \\(\\frac{1}{3}\\) Explanation: Question 5.<\/p>\n (A) 25 cm Explanation: Introduction to Trigonometry Class 10 MCQ Questions with Answers Question 1. The value of the expression [cosec (75\u00b0 + 0) – sec (15\u00b0 – 0) – tan (55\u00b0 + 0) + cot (35\u00b0 – 0)] is: (A) -1 (B) 0 (C) 1 (D) Answer: (B) 0 Explanation: cosec (75\u00b0 + 0) – sec (15\u00b0 – …<\/p>\n
\n\\(\\left[\\frac{\\sin ^{2} 22^{\\circ}+\\sin ^{2} 68^{\\circ}}{\\cos ^{2} 22^{\\circ}+\\cos ^{2} 68^{\\circ}}+\\sin ^{2} 63^{\\circ}+\\cos 63^{\\circ} \\sin 27^{\\circ}\\right]\\)
\n\\(=\\frac{\\sin ^{2} 22^{\\circ}+\\sin ^{2}\\left(90^{\\circ}-22^{\\circ}\\right)}{\\cos ^{2}\\left(90^{\\circ}-68^{\\circ}\\right)+\\cos ^{2} 68^{\\circ}}+\\sin ^{2} 63^{\\circ}\\) + cos 63\u00b0 sin (90\u00b0 – 63\u00b0)
\n= \\(\\frac{\\sin ^{2} 22^{\\circ}+\\cos ^{2} 22^{\\circ}}{\\cos ^{2} 68^{\\circ}+\\sin ^{2} 68^{\\circ}}+\\sin ^{2} 63^{\\circ}+\\cos 63^{\\circ} \\cdot \\cos 63^{\\circ}\\)
\n[\u2234 sin(90\u00b0 – \u03b8 ) = cos \u03b8 and cos (90\u00b0 – \u03b8 ) = sin \u03b8 ]
\n= \\(\\vec{a}\\) + (sin2<\/sup> 63\u00b0 + cos2<\/sup> 63\u00b0)
\n[\u2234 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1]
\n= 1 + 1 = 2<\/p>\n<\/p>\nIf 4 tan \u03b8 = 3, then \\(\\left(\\frac{4 \\sin \\theta-\\cos \\theta}{4 \\sin \\theta+\\cos \\theta}\\right)\\) is equal to:<\/h2>\n
\n(B) \\(\\frac{1}{2}\\)
\n(C) \\(\\frac{1}{3}\\)
\n(D) \\(\\frac{3}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{2}\\)<\/p>\n
\nGiven, 4 tan \u03b8 = 3
\n\u2234 tan \u03b8 = \\(\\frac{3}{4}\\)
\n\u2234 \\(\\frac{4 \\sin \\theta-\\cos \\theta}{4 \\sin \\theta+\\cos \\theta}\\) = \\(4 \\frac{4 \\frac{\\sin \\theta}{\\cos \\theta}-1}{4 \\frac{\\sin \\theta}{\\cos \\theta}+1}\\)
\n[Divided by cos \u03b8 in both numerator and denominator]
\n= \\(\\frac{4 \\tan \\theta-1}{4 \\tan \\theta+1}\\)[ \u2234 [tan\u03b8 = \\(\\frac{\\sin \\theta}{\\cos \\theta}\\)]
\n= \\(\\frac{4\\left(\\frac{3}{4}\\right)-1}{4\\left(\\frac{3}{4}\\right)+1}\\) = \\(\\frac{3-1}{3+1}\\) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\) [ Put tan \u2234 = \\(\\frac{3}{4}\\) frome equation (i)<\/p>\nIf cos A =\\(\\frac{4}{5}\\) then the value of tan A is:<\/h2>\n
\n(B) \\(\\frac{3}{4}\\)
\n(C) \\(\\frac{4}{3}\\)
\n(D) \\(\\frac{1}{8}\\)
\nAnswer:
\n(B) \\(\\frac{3}{4}\\)<\/p>\n
\nGiven,
\ncosA = \\(\\frac{4}{5}\\)
\n\u2234 sin A = \\(\\sqrt{1-\\cos ^{2} A}\\)
\n[[\u2234 sin2 A + cos2 A = 1 \u2234 sin A = \\(\\left.\\sqrt{1-\\cos ^{2} A}\\right]\\)
\nsin A = \\(\\sqrt{1-\\left(\\frac{4}{5}\\right)^{2}}\\) = \\(\\sqrt{1-\\frac{16}{25}}\\) = \\(\\sqrt{\\frac{9}{25}}\\) = \\(\\frac{3}{5}\\)
\ntan A = \\(\\frac{\\sin A}{\\cos A}\\)
\n= \\(\\frac{\\frac{3}{5}}{\\frac{4}{5}}\\) =\\(\\frac{3}{4}\\)<\/p>\nIf sin A = \\(\\frac{1}{2}\\) then the value of cot A is:<\/h2>\n
\n(B) \\(\\frac{1}{\\sqrt{3}}\\)
\n(C) \\(\\frac{\\sqrt{3}}{2}\\)
\n(D) 1
\nAnswer:
\n(A) \u221a 3<\/p>\n
\nGiven, sin A = \\(\\frac{1}{2}\\)
\ncos A = \\(\\sqrt{1-\\sin ^{2} A}\\) = \\(\\sqrt{1-\\left(\\frac{1}{2}\\right)^{2}}\\)
\ncos A = \\(\\sqrt{1-\\frac{1}{4}}\\) =\\(\\sqrt{\\frac{3}{4}}\\) = \\(\\frac{\\sqrt{3}}{2}\\)
\n[\u2234 sin2<\/sup> A + cos2<\/sup> A = 1 \u03b8 cos A = \\(\\sqrt{1-\\sin ^{2} A}\\)]
\nNow, cot A = \\(\\frac{\\cos A}{\\sin A}\\) = \\(\\frac{\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}\\) = \u03b8 3<\/p>\nGiven that sin \u03b8\u00a0 = \\(\\frac{a}{b}\\) then cos \u03b8 is equal to<\/h2>\n
\n(B) \\(\\frac{b}{a}\\)
\n(C) \\(\\frac{\\sqrt{b^{2}-a^{2}}}{b}\\)
\n(D) \\(\\frac{a}{\\sqrt{b^{2}-a^{2}}}\\)
\nAnswer:
\n(C) \\(\\frac{\\sqrt{b^{2}-a^{2}}}{b}\\)<\/p>\n
\nGiven, sin \u03b8 = \\(\\frac{a}{b}\\)
\n[ \u2234 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1 \u03b8 cos \u03b8 = \\([\\left.\\sqrt{1-\\sin ^{2} \\theta}\\right]\\)
\ncos \u03b8 = \\(\\sqrt{1-\\left(\\frac{a}{b}\\right)^{2}}\\) = \\(\\sqrt{1-\\frac{a^{2}}{b^{2}}}\\) = \\(\\frac{\\sqrt{b^{2}-a^{2}}}{b}\\)<\/p>\nIf sin A + sin2<\/sup> A = 1,then the value of the expression (cos2<\/sup> A + cos4<\/sup> A) is:<\/h2>\n
\n(B) \\(\\frac{1}{2}\\)
\n(C) 2
\n(D) 3
\nAnswer:
\n(A) 1<\/p>\n
\nGiven, sin A + sin2<\/sup> A = 1
\n\u2234 sin A = 1 – sin2<\/sup> A= cos2<\/sup>A
\n[\u2234 sin2 \u03b8 + cos2 \u03b8 = 1]
\nOn squaring both sides ,we get
\nsin2<\/sup> A = cos4<\/sup> A
\n\u2234 1 – cos2 A = cos4 A
\n\u2234 cos2<\/sup> A + cos4<\/sup> A = 1<\/p>\n<\/p>\nThe vale of 9 sec2<\/sup> A – 9 tan2<\/sup> A is<\/h2>\n
\n(B) 9
\n(C) 8
\n(D) 0
\nAnswer:
\n(B) 9<\/p>\n
\n9 sec2<\/sup> A – 9 tan2<\/sup> A = 9(sec2<\/sup> A – tan2<\/sup> A)
\n= 9 (1) [\u2234 sec2<\/sup> A – tan2<\/sup> A = 1]
\n= 9<\/p>\nThe value of (1 + tan \u03b8 + sec \u03b8 )( 1 + cot \u03b8 – cosec \u03b8 ) is<\/h2>\n
\n(B) 1
\n(C) 2
\n(D) -1
\nAnswer:<\/p>\n
\n(1 + tan \u03b8 + sec \u03b8 )( 1 + cot \u03b8 – cosec\u03b8 )
\n= \\(\\left(1+\\frac{\\sin \\theta}{\\cos \\theta}+\\frac{1}{\\cos \\theta}\\right)\\)\\(\\left(1+\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{1}{\\sin \\theta}\\right)\\)
\n= \\(\\left(\\frac{\\cos \\theta+\\sin \\theta+1}{\\cos \\theta}\\right)\\)\\(\\left(\\frac{\\sin \\theta+\\cos \\theta-1}{\\sin \\theta}\\right)\\)
\n= \\(\\frac{(\\sin \\theta+\\cos \\theta)^{2}-(1)^{2}}{\\sin \\theta \\cos \\theta}\\)
\n= \\(\\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta+2 \\sin \\theta \\cos \\theta-1}{\\sin \\theta \\cos \\theta}\\)
\n= \\(\\frac{1+2 \\sin \\theta \\cos \\theta-1}{\\sin \\theta \\cos \\theta}\\)
\n= \\(\\frac{2 \\sin \\theta \\cos \\theta}{\\sin \\theta \\cos \\theta}\\)
\n= 2<\/p>\nThe value of (sec A + tan A)(1 – sin A) is<\/h2>\n
\n(B) sin A
\n(C) cosec A
\n(D) cos A
\nAnswer:
\n(D) cos A<\/p>\n
\n(sec A + tan A) (1 – sin A)
\n= \\(\\left(\\frac{1}{\\cos A}+\\frac{\\sin A}{\\cos A}\\right)\\) (1 – sin A)
\n= \\(\\left(\\frac{1+\\sin A}{\\cos A}\\right)\\) (1 – sin A)
\n= \\(\\left(\\frac{1-\\sin ^{2} A}{\\cos A}\\right)\\) = \\(\\frac{\\cos ^{2} A}{\\cos A}\\)
\n= cos A<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): Cot A is the product of Cot and A.
\nReason (R): The value of sin0 increases as 0 increases.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\ncot A is not the product of cot and A. It is the cotangent of \u03b8 A.
\n\u2234 Assertion is incorrect.
\nIn case of reason:
\nThe value of sin 0 increases as 0 increases in interval of \u03b8 \u00b0< \u03b8 \u00b0 < 90\u00b0 \u2234 Reason is correct: Hence, assertion is incorrect but reason is correct. Question 2. Assertion (A): The value of \\(\\frac{\\tan 47^{\\circ}}{\\cot 43^{\\circ}}\\) is 1. Reason (R): The value of the expression (sin 80\u00b0 – cos 80\u00b0) is negative. Answer: (C) A is true but R is false Explanation: In case of assertion: = \\(\\frac{\\tan 47^{\\circ}}{\\cot 43^{\\circ}}\\) = \\(\\frac{\\tan 47^{\\circ}}{\\cot \\left(90^{\\circ}-47^{\\circ}\\right)}\\) = \\(\\frac{\\tan 47^{\\circ}}{\\tan 47^{\\circ}}\\) = 1 \u2234 Assertion is correct. In case of reason: 80\u00b0 is near to 90\u00b0, sin 90\u00b0 = 1 and cos 90\u00b0 = 0 So, the given expression sin 80\u00b0 – cos 80\u00b0 > 0 So, the value of the given expression is positive.
\n\u2234 Reason is incorrect:
\nHence, assertion is correct but reason is incorrect.<\/p>\nAssertion (A): If tan A = cot B, then the value of (A + B) is 90\u00b0.
\nReason (R): If sec \u03b8 sin \u03b8 = 0, then the value of \u03b8 is 0\u00b0.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\ntan A = cot B (Given)
\ntan A = tan(90\u00b0 – B)
\n[\u2234 tan (90\u00b0 – \u03b8 ) = cot \u03b8 ] A = 90\u00b0 – B
\nA + B =90\u00b0.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nGiven, sec \u03b8 .sin \u03b8 = 0
\n\\(\\vec{a}\\) = 0
\nor, tan \u03b8 = 0 = tan 0\u00b0
\n\u2234 \u03b8 = 0\u00b0
\n\u2234 Reason is correct:
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nAssertion (A): If x = 2 sin2<\/sup> \u03b8 and y = 2 cos2<\/sup>\u03b8 + 1 then the value of x + y = 3.
\nReason (R): If tan \u03b8 = \\(\\frac{5}{12}\\) , then the value of sec \u03b8 is \\(\\frac{13}{12}\\)<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nwe have x = 2 sin2<\/sup>\u03b8
\nand y = 2 cos2<\/sup> \u03b8 + 1
\nThen, x + y = 2 sin2 \u03b8 + 2cos2 + 1
\n= 2(sin2 \u03b8 + cos2 \u03b8 ) + 1
\n= 2 \u00d7 1 + 1[\u2234 sin2 \u03b8 + cos2 \u03b8 = 1]
\n= 2 + 1 = 3
\n\u2234 Assertion is correct.
\nIn case of reason:
\ntan \u03b8 = \\(\\frac{5}{12}\\)
\nUnsing identity; sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1
\nsec2<\/sup>\u03b8 = 1 + tan2<\/sup> \u03b8
\nsec2<\/sup> \u03b8 = 1 + \\(\\left(\\frac{5}{12}\\right)^{2}\\)
\n= 1 + \\(\\frac{25}{144}\\)
\n= \\(\\frac{144+25}{144}\\)
\n= \\(\\frac{169}{144}\\)
\n= \\(\\sqrt{\\frac{13^{2}}{12^{2}}}\\)
\n= \\(\\frac{13}{12}\\)
\n\u2234 Reason is correct: Hence, both assertion but reason is not the assertion.
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n<\/p>\nAssertion (A): If k + 1 = sec2<\/sup> \u03b8 (1 + sin \u03b8 ) (1 – sin \u03b8 ), then the value of k’ is 1.
\nReason (R): If sin \u03b8 + cos \u03b8 =\u03b8 3 then the value of tan \u03b8 + cot \u03b8 is 1.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nk + 1 = sec2<\/sup> \u03b8 (1 + sin \u03b8 )(1 – sin \u03b8 )
\nor, k + 1 = sec2<\/sup> \u03b8 (1 – sin2<\/sup>\u03b8 )
\nor, k + 1 = sec2<\/sup>\u03b8 .cos2<\/sup> \u03b8
\n[\u2234 sin 2 \u03b8 + cos2<\/sup> \u03b8 = 1]
\nor, k + 1 = sec2<\/sup>\u03b8 \u00d7 \\(\\frac{1}{\\sec ^{2} \\theta}\\)
\nor, k + 1 = 1
\nor, k = 1 – 1
\n\u2234 k = 0.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nGiven sin \u03b8 + cos\u03b8 = \u03b8 3
\nsin2 \u03b8 + cos2 \u03b8 + 2 sin \u03b8 cos \u03b8 = 3
\n1 + 2 sin \u03b8 cos \u03b8 = 3
\n2 sin \u03b8 cos \u03b8 = 2
\nsin \u03b8 cos \u03b8 = 1 …..(i)
\ntan \u03b8 + cot \u03b8 = \\(\\frac{\\sin \\theta}{\\cos \\theta}\\) + \\(\\frac{\\cos \\theta}{\\sin \\theta}\\)
\n= \\(\\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\cos \\theta \\sin \\theta}\\)
\n= \\(\\frac{1}{\\cos \\theta \\sin \\theta}\\)
\n= \\(\\frac{1}{1}\\) = 1 [From equation (i)]
\n\u2234 Reason is correct
\nHence, Assertion is incorrect but reason is correct.<\/p>\nAssertion (A): If sin A = \\(\\frac{\\sqrt{3}}{2}\\), then the value of 2 cot2<\/sup> A – 1 is \\(\\frac{-1}{3}\\)
\nReason (R) : If \u03b8 be an acute angle and 5 cosec \u03b8 = 7, then the value of sin \u03b8 + cos2<\/sup>\u03b8 – 1 is 10.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\n2 cot2<\/sup> A – 1 = 2(cosec2<\/sup> A – 1) – 1
\n(\u2234 cot 2<\/sup> \u03b8 = – 1 + cosec2<\/sup>\u03b8
\n= \\(\\frac{2}{\\sin ^{2} A}\\) – 3
\n= \\(\\frac{2}{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}}\\) – 3
\n\u2234 2cot2<\/sup> A – 1 =\\(\\frac{8}{3}\\) – 3 = \\(\\frac{-1}{3}\\)
\n\u2234 Assertion is correct.
\nIn case of reason:
\nGiven, 5 cosec \u03b8 = 7
\nor, cosec \u03b8 = \\(\\frac{7}{5}\\)
\nsin \u2234 = \\(\\frac{5}{7}\\)
\n[\u2234 cosec \u03b8 = \\(\\frac{1}{\\sin \\theta}\\)]
\nsin \u03b8 + cos2<\/sup> \u03b8 – 1 = sin \u03b8 -(1 – cos2<\/sup>\u03b8
\n= sin \u03b8 – sin2<\/sup>\u03b8
\n(\u2234 sin2<\/sup> \u03b8 + cos2<\/sup>\u2234 = 1)
\n= \\(\\frac{5}{7}\\) – \\(\\left(\\frac{5}{7}\\right)^{2}\\)
\n= \\(\\frac{35-25}{49}\\) = \\(\\frac{10}{49}\\)
\n\u2234 Reason is correct
\nHence, Assertion is incorrect but reason is correct.<\/p>\n
\nI. Read the following text and answer the following question on the basis of the same:
\n‘Skvsails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The skv sails technology
\nallows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Introduction-to-Trigonometry-2.png?resize=388%2C225&ssl=1\")
<\/p>\nIn the given figure, if tan \u03b8 = cot (30\u00b0 + \u03b8 ), where 0 and 30\u00b0 + \u03b8 are acute angles, then the value of 0 is:<\/h2>\n
\n(B) 30\u00b0
\n(C) 60\u00b0
\n(D) None of these
\nAnswer:
\n(B) 30\u00b0<\/p>\n
\nGiven,
\ntan \u03b8 = cot(30\u00b0 + \u03b8 )
\n= tan[90\u00b0 – (30\u00b0 + \u03b8 )]
\n= tan(90\u00b0 – 30\u00b0 – \u03b8 )
\n\u2234 tan \u03b8 = tan(60\u00b0 -\u03b8 )
\n\u2234 \u03b8 = 60\u00b0 – \u03b8
\n\u2234 2\u03b8 = 60\u00b0
\n\u2234 \u03b8 = 30\u00b0<\/p>\nThe value of tan 30\u00b0. cot 60\u00b0 is:<\/h2>\n
\n(B) \\(\\frac{1}{\\sqrt{3}}\\)
\n(C) 1
\n(D) \\(\\frac{1}{3}\\)
\nAnswer:
\n(D) \\(\\frac{1}{3}\\)<\/p>\n
\ntan 30\u00b0 x cot 60\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\) \u00d7 \\(\\frac{1}{\\sqrt{3}}\\)
\n= \\(\\frac{1}{3}\\)<\/p>\nWhat should be the length of the rope of the kite sail in order to pull the ship at the angle 0 and be at a vertical height of 200 m<\/h2>\n
\n(B) 300 m
\n(C) 100 m
\n(D) 200 m
\nAnswer:
\n(A) 400 m<\/p>\n
\nIn \u2206ABC, we have 0 = 30\u00b0, AB = 200 m
\nThen, sin 30\u00b0 = \\(\\frac{\\text { Perpendicular}}{\\text { Hypotenuse}}\\)
\n= \\(\\frac{A B}{A C}\\)
\n\u2234 \\(\\frac{1}{2}\\) = \\(\\frac{200}{A C}\\)
\n\u2234 AC = 400 m.<\/p>\n<\/p>\nIf cos A = \\(\\frac{1}{2}\\), then the value of 9 cot2<\/sup> A – 1 is:<\/h2>\n
\n(B) 3
\n(C) 2
\n(D) 4
\nAnswer:
\n(C) 2<\/p>\n
\nGiven, cos A = \\(\\frac{1}{2}\\)
\n\u2234 cos A = cos 60\u00b0
\n\u2234 A = 60\u00b0
\nthen, 9 cot2<\/sup> A – 1 = 9(cot 60\u00b0)2<\/sup> – 1
\n= 9 \\(\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\) – 1
\n= 9 \u00d7 \\(\\frac{1}{3}\\) – 1 = 3 – 1
\n= 2<\/p>\nIn the given figure, the value of (sin C + cos A) is:<\/h2>\n
\n(B) 2
\n(C) 3
\n(D) 4
\nAnswer:
\n(A) 1<\/p>\n
\nWe have,
\nAB = 200 m and AC = 400 m Proved in Question 3]
\nThen, sin C + cos A = \\(\\frac{A B}{A C}\\) + \\(\\frac{A B}{A C}\\)
\n= 2 \u00d7 \\(\\frac{A B}{A C}\\)
\n= 2 \u00d7 \\(\\frac{200}{400}\\) = 1<\/p>\n
\nAuthority wants to construct a slide in a city park for children. The slide was to be constructed for children below the age of 12 years. Authority prefers the top of the slide at a height of 4 m above the ground and inclined at an angle of 30\u00b0 to the ground.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Introduction-to-Trigonometry-3.png?resize=303%2C196&ssl=1\")
<\/p>\nThe distance of AB is:<\/h2>\n
\n(B) 6 m
\n(C) 5 m
\n(D) 10 m
\nAnswer:
\n(A) 8 m<\/p>\n
\n\u2220B = 30\u00b0 and
\nAC = 4 m
\nThen, sin 30\u00b0 = \\(\\frac{A C}{A R}\\)
\n\u21d2 \\(\\frac{1}{2}\\) = \\(\\frac{4}{A B}\\)
\n\u21d2 AB = 8 m.<\/p>\nIn value of sin 2<\/sup> 30\u00b0 + cos 2<\/sup> 60\u00b0 is:<\/h2>\n
\n(B) \\(\\frac{1}{2}\\)
\n(C) \\(\\frac{3}{4}\\)
\n(D) \\(\\frac{3}{2}\\)
\nAnswer:
\n(B) \\(\\frac{1}{2}\\)
\nsin 2<\/sup> 30\u00b0 + cos2<\/sup> 60\u00b0= \\(\\left(\\frac{1}{2}\\right)^{2}\\) + \\(\\left(\\frac{1}{2}\\right)^{2}\\)
\n= \\(\\frac{1}{4}\\) = \\(\\\\frac{1}{4}vec{a}\\).
\n= \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\)<\/p>\nIf cos A = \\(\\vec{a}\\), then the value of cot2<\/sup>A – 2 is:<\/h2>\n
\n(B) 4
\n(C) 3
\n(D) 2
\nAns:
\n(D) 2<\/p>\n
\nsince, cos A = \\(\\vec{a}\\)
\n\u21d2 cos A = cos 60\u00b0
\n\u21d2 A = 60\u00b0
\nThen 12 cot2<\/sup> A -2 = 12(cot 60\u00b0) – 2
\n= 12 \\(\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\) – 2
\n= 12 \u00d7 \\(\\frac{1}{3}\\) – 2
\n= 4 – 2 = 2.<\/p>\nIn the given figure, the value of (sin C \u00d7 cos A ) is:<\/h2>\n
\n(B) \\(\\frac{1}{2}\\)
\n(C) \\(\\frac{1}{4}\\)
\n(D) \\(\\frac{1}{5}\\)
\nAnswer:
\n(B) \\(\\frac{1}{2}\\)<\/p>\n
\nsince, AC \u22a5BC,
\nthen \u2220C = 90\u00b0
\nsin C cos A = sin 90 \u00d7 \\(\\frac{A C}{A B}\\)
\n= 1 \u00d7 \\(\\frac{4}{8}\\)
\n= \\(\\frac{1}{2}\\)<\/p>\n<\/p>\nIn the given figure, if AB + BC = 258 cm and AC = 5 cm, then the value of BC is:<\/h2>\n
\n(B) 15 cm
\n(C) 10 cm
\n(D) 12 cm
\nAnswer:
\n(D) 12 cm<\/p>\n
\nwe have, \u2220C = 90
\nAB = BC = 25 cm and AC = 5 cm
\nlet BC be x cm, then AB = (25 – x ) cmBy using Pythagoras theorem ,
\nAB2<\/sup> = BC 2<\/sup>+ AC2<\/sup>
\n\u21d2 (25 – x2<\/sup>) = x2<\/sup> + (5)2<\/sup>
\n\u21d2 625 – 50 x + x2<\/sup> + 25
\n\u21d2 50x = 600
\n\u21d2 x = \\(\\frac{600}{50}\\) = 12
\nHence, BC = 12 cm<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"