If sinA , cos A and tanA are in geometric progression then cot ^6 A - cot ^2 A is ___________
a) 2 b) 3 c)4 d) 1
explain
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Answered by
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If sinA, cosA and tanA are in GP, then
CosA/SinA = tanA/cosA
Cos^2A = tanA.sinA (Cross-multiplying)
Cos^3A = sin^2A
(Simplifying using tanA = sinA/cosA)------------(1)
CosA/SinA = tanA/cosA (Given)
CotA = 1/cotA .1/cosA
Cot^2A = 1/cosA = secA (Cross-multiplying)-------(2)
Then, cot^6A = sec^3A (On cubing both sides of 2)----------(3)
Now, equation(3) -equation(2)
= cot^6A - cot^2A
= sec^3A - secA
=secA(sec^2A - 1) [taking secA common term outside]
=secA.tan^2A [from
Identity sec^2A - tan^2A = 1]
=1/cosA . Sin^2A/cos^2A
=sin^2A/cos^3A..............4
All the best
Substituting 1 in 4
We know, sin^2A = cos^3A
So,
Cot^6A - Cot^2A = 1
CosA/SinA = tanA/cosA
Cos^2A = tanA.sinA (Cross-multiplying)
Cos^3A = sin^2A
(Simplifying using tanA = sinA/cosA)------------(1)
CosA/SinA = tanA/cosA (Given)
CotA = 1/cotA .1/cosA
Cot^2A = 1/cosA = secA (Cross-multiplying)-------(2)
Then, cot^6A = sec^3A (On cubing both sides of 2)----------(3)
Now, equation(3) -equation(2)
= cot^6A - cot^2A
= sec^3A - secA
=secA(sec^2A - 1) [taking secA common term outside]
=secA.tan^2A [from
Identity sec^2A - tan^2A = 1]
=1/cosA . Sin^2A/cos^2A
=sin^2A/cos^3A..............4
All the best
Substituting 1 in 4
We know, sin^2A = cos^3A
So,
Cot^6A - Cot^2A = 1
mysticd:
ur solution is correct , but it is too lengthy . we can do it in short also
yes
Answered by
1
Answer:
If sinA, cosA and tanA are in GP, then
CosA/SinA = tanA/cosA
Cos^2A = tanA.sinA (Cross-multiplying)
Cos^3A = sin^2A
(Simplifying using tanA = sinA/cosA)------------(1)
CosA/SinA = tanA/cosA (Given)
CotA = 1/cotA .1/cosA
Cot^2A = 1/cosA = secA (Cross-multiplying)-------(2)
Then, cot^6A = sec^3A (On cubing both sides of 2)----------(3)
Now, equation(3) -equation(2)
= cot^6A - cot^2A
= sec^3A - secA
=secA(sec^2A - 1) [taking secA common term outside]
=secA.tan^2A [from
Identity sec^2A - tan^2A = 1]
=1/cosA . Sin^2A/cos^2A
=sin^2A/cos^3A..............4
Substituting 1 in 4
We know, sin^2A = cos^3A
Cot^6A - Cot^2A = 1
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