If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tan α + tan β = 2b/(a + c).
Note
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Answers
Answer:
Step-by-step explanation:
Answer: (2) 2b/(c + a)
Solution:
Given,
a cos 2θ + b sin 2θ = c….(i)
a[(1 – tan2θ)/(1 + tan2θ)] + b[2 tan θ/(1 + tan2θ)] = c
a(1 – tan2θ) + 2b tan θ = c(1 + tan2θ)
c + c tan2θ – a + a tan2θ – 2b tan θ = 0
(c + a)tan2θ – 2b tan θ + (c – a) = 0….(ii)
Given that α and β are the roots of equation (i).
Thus, tan α and tan β are the roots of equation (ii) as it is a quadratic equation in tan θ.
Sum of the roots of equation (ii),
tan α + tan β = -(-2b)/(c + a) = 2b/(c + a)
the sum of the lower limits of the median and modal class is
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FOLLOW
Given that,
Now, we have the equation,
We know that,
and
So,
On substituting the values, in given equation, we get
On taking LCM, we get
can be rewritten as
Now, its a quadratic in tan,
So,
We know that,
Thus,
Hence, Proved