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=a^2×2=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 a^4=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.


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