## MATHEMATICS TRIGONOMETRY RATIOS AND IDENTITIES QUIZ-4

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1.

•  Right angled
•  Obtuse angled
•  Equilateral
•  Isosceles
Solution

Q2.If 5cos⁡2Î¸+2 cos2Î¸/2 +1=0,-Ï€ < Î¸ < Ï€, then Î¸ is equal to
•  Ï€/3
•  Ï€/3,cos-1(3/5)
•  cos-1(3/5)
•  Ï€/3,Ï€-cos-1⁡(3/5)
Solution

Q3.  If cos(Î±+Î²)=4/5, sin (Î±-Î²)=5/13 and Î±,Î² lies between 0 and Ï€/4, then tan 2Î± is equal to
•  16/63
•  56/33
•  28/33
•  None of these
Solution

Q4.

•  1
•  2
•  3
•  4
Solution

Q5.If sin⁡x+cos⁡x=1/5, then tan⁡2x is
•  25/17
•  7/25
•  25/7
Solution

Q6. If the altitudes of a triangle are in AP, then the sides of the triangle are in
•  A.P
•  G.P
•  H.P
•  None of these
Solution

Q7.If A=35°,B=15° and C=40°, then tan⁡A.tan⁡B+tan⁡B.tan⁡C+tan⁡C .tan⁡A is equal to
• 0
•  1
•  2
•  3
Solution

Q8.If sin⁡x cos⁡x cos⁡2x=Î» has a solution, then Î» lies in the interval
• [-1/4,1/4]
•  [-1/2,1/2]
•  (-∞,-1/4]∪[1/4,∞)
•  (-∞,-1/2]∪[1/2,∞)
Solution

Q9.If f(x)=cos2⁡x+sec2⁡x, its value always is
•  f(x)< 1
•  f(x)=1
•  2>f(x)>1
•  f(x)≥2
Solution

Q10. The value of tan⁡〖Î±+2 tan⁡(2Î±)+4 tan⁡(4Î±)+...+2n-1tan⁡(2n-1 Î±)+2n cot⁡(2n Î±) is
•  cot⁡(2nÎ±)
•  2n tan⁡(2nÎ±)
•  0
•  cot⁡Î±
Solution

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