Practice Questions 2-Introduction to Trigonometry-Class X
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Question 1 of 30
1. Question
1 point(s)The value of \(\)\dfrac{{({{\sin }^3}Q + {{\cos }^3}Q)}}{{\sin Q + \cos Q}}[\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{{{{\sin }^3}Q + {{\cos }^3}Q}}{{\sin Q + \cos Q}} = \dfrac{{(\sin Q + \cos Q)({{\sin }^2}Q – \sin Q\cos Q + {{\cos }^2}Q)}}{{\sin Q + \cos Q}}\\ = {\sin ^2}Q + {\cos ^2}Q – \sin Q + \cos Q\\ = 1 – \sin Q\cos Q\end{array}[\latex]
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Question 2 of 30
2. Question
1 point(s)The value of \(\){(\dfrac{{\sin 27}}{{\cos 63}})^{^2}} + {(\dfrac{{\cos 63}}{{\sin 27}})^2}[\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}{[\dfrac{{\sin (90 – 63)}}{{\cos 63}}]^2} + {[\dfrac{{\cos (90 – 27)}}{{\sin 27}}]^2}\\ = {(\dfrac{{\cos 63}}{{\cos 63}})^2} + {(\dfrac{{\sin 27}}{{\sin 27}})^2}\\ = 1 + 1 = 2\end{array}[\latex]
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Question 3 of 30
3. Question
1 point(s)The value of \(\)\dfrac{{\cos (90 – Q)}}{{\sin Q}} + \dfrac{{\sin Q}}{{\cos (90 – Q)}}[\latex]
CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{{\cos (90 – Q)}}{{\sin Q}} + \dfrac{{\sin Q}}{{\cos (90 – Q)}}\\ = \dfrac{{\sin Q}}{{\sin Q}} + \dfrac{{\sin Q}}{{\sin Q}} = 1 + 1 = 2\end{array}[\latex]
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Question 4 of 30
4. Question
1 point(s)The value of \(\)\dfrac{{\cos (90 – Q)\cos Q}}{{\tan Q}} + {\cos ^2}(90 – Q)[\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{{\cos (90 – Q)\cos Q}}{{\tan Q}} + {\cos ^2}(90 – Q)\\ = \dfrac{{\sin Q\cos Q}}{{\tan Q}} + {\sin ^2}Q\\ = {\cos ^2}Q + {\sin ^2}Q\\ = 1\end{array}[\latex]
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Question 5 of 30
5. Question
1 point(s)The value of \(\){\sec ^2}42 – \cos e{c^2}48[\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}{\sec ^2}42 – \cos e{c^2}(90 – 42)\\{\sec ^2}42 – {\sec ^2}42\\ = 0\end{array}[\latex]
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Question 6 of 30
6. Question
1 point(s)If (1+cosA)(1-cosA)=\(\)\dfrac{3}{4}[\latex],the value of secA is
CorrectIncorrectHint
\(\)\begin{array}{l}(1 + \cos A)(1 – \cos A) = \dfrac{3}{4}\\1 – {\cos ^2}A = \dfrac{3}{4}\\1 – \dfrac{3}{4} = {\cos ^2}A\\{\cos ^2}A = \dfrac{1}{4}\\{\sec ^2}A = \dfrac{1}{4}\\\sec A = \pm 2\end{array}[\latex]
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Question 7 of 30
7. Question
1 point(s)If sin(A+B)=1and tan(A-B)=\(\)\dfrac{1}{{\sqrt 3 }}[\latex]the value of tanA+cotB is
CorrectIncorrectHint
\(\)\begin{array}{l}\sin (A + B) = 1\\ = A + B = 90\_\_\_\_1\\\tan (A – B) = \dfrac{1}{{\sqrt 3 }}\\ = A – B = 30\_\_\_\_\_\_2\end{array}[\latex]
Solving both the equations we get A=60,B=30
TanA+cotB=tan60+cot30=
\(\)\sqrt 3 + \sqrt 3 = 2\sqrt 3 [\latex] -
Question 8 of 30
8. Question
1 point(s)If sin Q=x, sec Q=x ,sec Q=the value of cot Q is
CorrectIncorrectHint
\(\)\begin{array}{l}\cot Q = \dfrac{{\cos Q}}{{\sin Q}}\\\sec Q = Y\\\cos Q = \dfrac{1}{Y}\\\cot Q = \dfrac{1}{{XY}}\end{array}[\latex]
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Question 9 of 30
9. Question
1 point(s)If cossecQ=\(\)\dfrac{5}{3}[\latex], the value of cosQ+tanQ is
CorrectIncorrectHint
\(\)\begin{array}{l}BC = \sqrt {25{k^2} – 9{k^2}} \\ = \sqrt {16{k^2}} \\ = 4k\\\cos Q + \tan Q = \dfrac{4}{5} + \dfrac{3}{4}\\ = \dfrac{{16 + 15}}{{20}} = \dfrac{{31}}{{20}}\end{array}[\latex]
\(\)[\latex] -
Question 10 of 30
10. Question
1 point(s)The value of sin38-cos52 is
CorrectIncorrectHint
Sin38-cos52
=cos(90-38)-cos52=cos52-cos52
=0 -
Question 11 of 30
11. Question
1 point(s)If cosQ=\(\)\dfrac{1}{2}[\latex],the value of cosQ+secQ is
CorrectIncorrectHint
\(\)\begin{array}{l}\cos Q = \dfrac{1}{2}\\ = \sec Q = 2\\\cos Q + \sec Q = \dfrac{1}{2} + 2 = \dfrac{{1 + 4}}{2} = \dfrac{5}{2}\end{array}[\latex]
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Question 12 of 30
12. Question
1 point(s)The value of \(\)3{\cot ^2} + {\sec ^2}45[\latex] is
CorrectIncorrectHint
\(\)3{(\dfrac{1}{{\sqrt 3 }})^2} + {(\sqrt 2 )^2} = 3 \times \dfrac{1}{2} + 2 = 1 + 2 = 3[\latex]
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Question 13 of 30
13. Question
1 point(s)If sinA=cosA,the value of \(\)2{\tan ^2}A + {\sin ^2} + 1[\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}\sin A = \cos A\\A = 45\\2{\tan ^2}45 + {\sin ^2}45 + 1\\ = 2 \times {1^2} + (\dfrac{1}{{\sqrt 2 }}) + 1\\ = 2 + \dfrac{1}{2} + 1 = \dfrac{{4 + 1 + 2}}{2} = \dfrac{7}{2}\end{array}[\latex]
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Question 14 of 30
14. Question
1 point(s)If tanQ+cotQ=2, the value of \(\)\sqrt {{{\tan }^2}Q + {{\cot }^2}Q} [\latex] is
CorrectIncorrectHint
\(\)\begin{array}{l}\tan Q + \cot Q = 2\\{\tan ^2}Q + {\cot ^2}Q + 2 = 4\\{\tan ^2}Q + {\cot ^2}Q = 2\\\sqrt {{{\tan }^2}Q + {{\cot }^2}Q} = \sqrt 2 \end{array}[\latex]
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Question 15 of 30
15. Question
1 point(s)If tan(A-B) and tan(A+B)=\(\)\sqrt 3 [\latex]
CorrectIncorrectHint
\(\)\begin{array}{l}\tan (A – B) = \dfrac{1}{{\sqrt 3 }}\\ = A – B = 30\_\_\_1\\\tan (A + B) = \sqrt 3 \\ = A + D = 60\_\_\_2\end{array}[\latex]
Solving both the equations we get
A=45,B=15 -
Question 16 of 30
16. Question
1 point(s)The value of \(\)\dfrac{{3\tan 25.\tan 40.\tan 50.\tan 65 – \dfrac{1}{2}{{\tan }^2}60}}{{4({{\cos }^2}29 + {{\cos }^2}61)}}[\latex]
CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{{3\tan 25\tan 40\tan (90 – 40)\tan (90 – 25) – \dfrac{1}{2} \times {{(\sqrt 3 )}^2}}}{{4[{{\cos }^2}29 + {{\cos }^2}(90 – 29)]}}\\\dfrac{{3\tan 25\tan 40\tan \cot 40\cot 25{{(\sqrt 3 )}^2}}}{{4[{{\cos }^2}29 + {{\sin }^2}29)]}}\\\dfrac{{3 – \dfrac{3}{2}}}{4} = \dfrac{{\dfrac{3}{2}}}{4} = \dfrac{3}{8}\end{array}[\latex]
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Question 17 of 30
17. Question
1 point(s)The value of
\(\)\dfrac{{4{{\cot }^2}60 + {{\sec }^2}30 – 2{{\sin }^2}45}}{{{{\sin }^2}60 + {{\cos }^2}45}}[\latex]CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{{4 \times {{(\dfrac{1}{{\sqrt 3 }})}^2} + {{(\dfrac{2}{{\sqrt 3 }})}^2} – 2 \times {{(\dfrac{1}{{\sqrt 2 }})}^2}}}{{{{(\dfrac{{\sqrt 3 }}{2})}^2} + {{(\dfrac{1}{{\sqrt 2 }})}^2}}}\\\dfrac{{\dfrac{4}{3} + \dfrac{4}{3} – 1}}{{\dfrac{3}{4} + \dfrac{1}{2}}} = \dfrac{{\dfrac{8}{3} – 1}}{{\dfrac{{3 + 2}}{4}}} = \dfrac{5}{3} \times \dfrac{4}{5} = \dfrac{4}{3}\end{array}[\latex]
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Question 18 of 30
18. Question
1 point(s)If sec + tanQ=P, the value of cosecQ is
CorrectIncorrectHint
\(\)\begin{array}{l}\sec Q + \tan Q = ?\_\_\_\_1\\{\sec ^2}Q – {\tan ^2}Q = 1\\(\sec Q + \tan Q)(\sec Q – \tan Q) = 1\\P(\sec Q – \tan Q) = 1\\\sec Q – \tan Q = \dfrac{1}{P}\_\_\_\_\_2\\Adding1and2\\2\sec Q = P + \dfrac{1}{2}\\\sec Q = \dfrac{{{P^2} + 1}}{{2P}}\\Subtracting1and2\\weget,\\2\tan Q = p – \dfrac{1}{p} = \tan Q = \dfrac{{{p^2} – 1}}{{2p}}\\\dfrac{{\sec Q}}{{\tan Q}} = \dfrac{{\dfrac{{{P^2} – 1}}{{2p}}}}{{\dfrac{{{p^2} – 1}}{{2p}}}} = \dfrac{{\dfrac{1}{{\cos Q}}}}{{\dfrac{{\sin Q}}{{\cos Q}}}} = \dfrac{{{P^2} + 1}}{{{P^2} – 1}}\\ = \dfrac{1}{{\sin Q}} = \dfrac{{{P^2} + 1}}{{{P^2} – 1}}\\ = \cos ecQ = \dfrac{{{P^2} + 1}}{{{P^2} – 1}}\end{array}[\latex]
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Question 19 of 30
19. Question
1 point(s)The value of \(\)\dfrac{4}{{{{\cot }^2}30}} + \dfrac{1}{{{{\sin }^2}60}} – {\cos ^2}45[\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{4}{{(\sqrt {3)} }} + \dfrac{1}{{{{(\dfrac{1}{{\sqrt 3 }})}^2}}} – {(\dfrac{1}{{\sqrt 2 }})^{^2}}\\ = \dfrac{4}{3} + \dfrac{4}{3} – \dfrac{1}{2} = \dfrac{{13}}{6}\end{array}[\latex]
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Question 20 of 30
20. Question
1 point(s)The value of \(\)4({\sin ^4}30 + {\cos ^4}60) – 3({\cos ^2}45 – {\sin ^2}90)[\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}4[{(\dfrac{1}{2})^4} + {(\dfrac{1}{2})^4}] – 3[{(\dfrac{1}{{\sqrt 2 }})^2} – 1]\\ = 4[\dfrac{1}{{16}} + \dfrac{1}{{16}}] – 3[\dfrac{1}{2} – 1]\\ = (4 \times \dfrac{2}{{16}}) – 3( – \dfrac{1}{2})\\ = \dfrac{1}{2} + \dfrac{3}{2} = 2\end{array}[\latex]
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Question 21 of 30
21. Question
1 point(s)The value of 2tan\(\)2{\tan ^3}45 + \cos [\latex]is
CorrectIncorrectHint
\(\)\begin{array}{l}2 \times {(1)^2}{(\dfrac{{\sqrt 3 }}{2})^2} – {(\dfrac{{\sqrt 3 }}{2})^2}\\ = 2 + 0 = 2\end{array}[\latex]
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Question 22 of 30
22. Question
1 point(s)The value of tan45 tan23tan42tan67 is
CorrectIncorrectHint
\(\)\begin{array}{l}\tan (90 – 42)\tan (90 – 67)\tan 42\tan 67\\ = \cot 42\cot 67\tan 42\tan 67\\ = (\cot 42\tan 42)(\cot 67\tan 67)\\ = 1 \times 1 = 1\end{array}[\latex]
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Question 23 of 30
23. Question
1 point(s)If tan2A=cot (A-18), where 2A is an acute angle. The value of A is
CorrectIncorrectHint
\(\)\begin{array}{l}\tan 2A = \cot (A – 18)\\Also,\\\tan 2A = \cot (90 – 2A)\\So,\\\cot (90 – 2A) = \cot (A – 18)\\90 – 2A = A – 18\\A = \dfrac{{108}}{3} = 36\end{array}[\latex]
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Question 24 of 30
24. Question
1 point(s)The value of \(\)\dfrac{{{{\sec }^2}Q – 1}}{{{{\tan }^2}Q}}[\latex] is
CorrectIncorrectHint
\(\)\dfrac{{{{\sec }^2} – 1}}{{{{\tan }^2}Q}} = \dfrac{{{{\tan }^2}Q}}{{{{\tan }^2}Q}} = 1[\latex]
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Question 25 of 30
25. Question
1 point(s)The value of \(\)\sqrt {\dfrac{{1 + \cos Q}}{{1 – \cos Q}}} + \sqrt {\dfrac{{1 – \cos Q}}{{1 + \cos Q}}} [\latex] is
CorrectIncorrectHint
\(\)\begin{array}{l}\sqrt {\dfrac{{(1 + \cos Q)(1 + \cos Q)}}{{(1 – \cos Q)(1 + \cos Q)}}} + \sqrt {\dfrac{{(1 – \cos Q)(1 – \cos Q)}}{{(1 + \cos Q)(1 – \cos Q)}}} \\ = \sqrt {\dfrac{{{{(1 + \cos Q)}^2}}}{{1 – {{\cos }^2}}}} + \sqrt {\dfrac{{{{(1 – \cos Q)}^2}}}{{1 – {{\cos }^2}Q}}} \\ = \dfrac{{1 + \cos Q}}{{\sin Q}} + \dfrac{{1 – \cos Q}}{{{\mathop{\rm Sin}\nolimits} Q}}\\ = \dfrac{{1 + \cos Q + 1 – \cos Q}}{{\sin Q}}\\ = 2\cos ecQ\end{array}[\latex]
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Question 26 of 30
26. Question
1 point(s)If \(\)\tan Q = \dfrac{a}{b}[\latex], the value of \(\)\dfrac{{a\sin Q – b\cos Q}}{{a\sin Q + b\cos Q}}[\latex] is
CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{{a\dfrac{{\sin Q}}{{\cos Q}} – b\dfrac{{\cos Q}}{{\cos Q}}}}{{a\dfrac{{\sin Q}}{{\cos Q}} + b\dfrac{{\cos Q}}{{\cos Q}}}} = \dfrac{{a \times \dfrac{a}{b} – b}}{{a \times \dfrac{a}{b} + b}}\\ = \dfrac{{\dfrac{{{a^2}}}{b} – b}}{{\dfrac{{{a^2}}}{b} + b}}\\ = \dfrac{{{a^2} – {b^2}}}{{{a^2} + {b^2}}}\end{array}[\latex]
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Question 27 of 30
27. Question
1 point(s)If \(\)\tan Q = \dfrac{X}{Y}[\latex], the value of sinQ is
CorrectIncorrectHint
\(\)\begin{array}{l}\tan Q = \dfrac{X}{Y} \Rightarrow CotQ = \dfrac{y}{x}\\\cos ecQ = \sqrt {1 + (\dfrac{Y}{X}} {)^2} = \dfrac{{\sqrt {{X^2} + {Y^2}} }}{X}\\So,\sin Q = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}\end{array}[\latex]
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Question 28 of 30
28. Question
1 point(s)If \(\)\sin A + {\sin ^2}A = 1[\latex], the value of \(\){\cos ^2}A + {\cos ^4}A[\latex] is
CorrectIncorrectHint
\(\)\begin{array}{l}\sin A = 1 – {\sin ^2}A = {\cos ^2}A\\{\cos ^2}A + {\cos ^4}A = {\cos ^2}A + {\sin ^2}A\\ = 1\end{array}[\latex]
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Question 29 of 30
29. Question
1 point(s)If \(\)\dfrac{{\sin Q + \cos Q}}{{\sin Q – \cos Q}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}}[\latex], the value of acute angle Q is
CorrectIncorrectHint
\(\)\begin{array}{l}\dfrac{{1 + \dfrac{{\cos Q}}{{\sin Q}}}}{{1 – \dfrac{{\cos Q}}{{\sin Q}}}} = \dfrac{{1 + \dfrac{1}{{\sqrt 3 }}}}{{1 – \dfrac{1}{{\sqrt 3 }}}}\\ = \dfrac{{1 + \cot Q}}{{1 – \cot Q}} = \dfrac{{1 + \dfrac{1}{{\sqrt 3 }}}}{{1 – \dfrac{1}{{\sqrt 3 }}}}\\ = \cot Q = \dfrac{1}{{\sqrt 3 }}\\\cot = 60\\Q = 60\end{array}[\latex]
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Question 30 of 30
30. Question
1 point(s)\(\)\dfrac{1}{{1 + \sin Q}} + \dfrac{1}{{1 – \sin Q}}[\latex]=
CorrectIncorrectHint
\(\)\dfrac{{1 – \sin Q + 1 + \sin Q}}{{1 – {{\sin }^2}Q}} = \dfrac{2}{{{{\cos }^2}Q}} = 2{\sec ^2}Q[\latex]
