10 Jan 2016

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Solved Problems on Trigonometry

SOLVED PROBLEMS ON TRIGONOMETRY:

Q. Find the value of sin2a+[1/(1+tan2a)] ?

A.   sec2a - tan2a=1.
=> sec2a=1+tan2a.
=> 1/sec2a=cos2a.
therefore, sin2a+cos2a = 1.
Hence, our Answer is 1.

Q. if 3cos2A+7sin2A=4 then find value of cotA, given that A is an acute angle?

A.  3cos2A+7sin2A=4, _______(1)
=>  cos2A+sin2A=1,
Therefore, 3cos2A+3sin2A=3, ________(2)
Putting eq(2) in eq(1),
We get  4sin2A=1,
Therefore sinA=1/2.=> A=30 degree,
Therefore cot30 = 1/root 3.


Q. if cos2a+cos4a=1, what is the value of tan2a+tan4a?

A.  sin2a+cos2a=1 ..............eq(1)

In the question, we are given that

cos2a+cos4a=1

taking cos2a on the right hand side

cos4a=1-cos2a

taking value from eq(1)

cos4a=sin2a

cos2a x cos2a = sin2a ; laws of surds and indices

cos2a=(sin2a/cos2a)

cos2a=tan2a…….............eq(2)

now lets move to the next part. in the question, we have to find the value of

tan2a+tan4a

=tan2a+(tan2a)2 ; because (a2)2=a2×2=4

= cos2a+cos4a ;applying value from eq2

=1 ;this was already given in the first part of the question itself.

Final answer=1.

Q.find value of sin4a-cos4a

We know that

(a2-b2)=(a+b)(a-b)

So instead of sin4a-cos4a, I can write

(sin2a)2-(cos2a)2 ; because a4=2×2 =(a2)2

=(Sin2a+cos2a)(Sin2a-cos2a)

=1 x (Sin2a-cos2a) ; because (a2-b2)=(a+b)(a-b)

=(Sin2a-cos2a)

Q. if sin4a-cos4a=-2/3 then what is the value of 2cos2a-1?

A. sin4a-cos4a=-2/3,
a2-b2=(a+b)(a-b)...........eq(1)
From eq(1),
sin4a-cos4a=(sin2-cos2)(sin2+cos2)
and from identities, sin2+cos2=1,
sin2-cos2=-2/3,
2cos2a-1=2/3.


CRACK SSC CGL 2016
LINKS
||Basics of Trigonometry||Unsolved Problems on Trigonometry||

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