please help,i don't understand these kind of problems. show the steps and explain. Thank you!
Unless you can see any nice method to do this, the best way to go is to convert everything in terms of sines and cosines.
Using tanθ = sinθ/cosθ and cotθ = sinθ/cosθ, the left side becomes:
LHS = (tanθ - cotθ)/(tanθ + cotθ) + 1
= (sinθ/cosθ - cosθ/sinθ)/(sinθ/cosθ + cosθ/sinθ) + 1
= (sin^2θ - cos^2θ)/(sin^2θ + cos^2θ) + 1, by multiplying top/bottom by sinθcosθ
= sin^2θ - cos^2θ + 1, since sin^2θ + cos^2θ = 1
= sin^2θ + (1 - cos^2θ), by re-arranging
= sin^2θ + sin^2θ, since sin^2θ + cos^2θ = 1
= 2sin^2θ
= RHS.
I hope this helps!
(tanΘ-cotΘ)/(tanΘ+cotΘ) + 1 = 2sin²Θ
LHS =
{(tanΘ-cotΘ)/(tanΘ+cotΘ)} +1
{(tanΘ-cotΘ)(tanΘ-cotΘ)/(tanΘ+cotΘ)(tanΘ-cotΘ)} +1
{(tanΘ-cotΘ)^2 / (tan^2Θ - cot^2Θ)} +1
{(tan^2Θ + cot^2Θ - 2tanΘcotΘ) / (tan^2Θ - cot^2Θ)} +1
* note : tanΘ.cotΘ = 1
{(tan^2Θ + cot^2Θ - 2) / (tan^2Θ - cot^2Θ)} +1
*note : taking (tan^2Θ - cot^2Θ) as the common denominator
(tan^2Θ + cot^2Θ - 2 + tan^2Θ - cot^2Θ) / (tan^2Θ - cot^2Θ)
(2tan^2Θ - 2) / (tan^2Θ - cot^2Θ)
*note : tan^2Θ +1 = sec^2Θ and cot^2Θ +1 = cosec^2Θ
(2sec^2Θ - 4) / (sec^2Θ - cosec^2Θ)
*note : sec^2Θ = 1/cos^2Θ and cosec^2Θ= 1/sin^2Θ
2([1/cos^2Θ] - 2) / ([1/cos^2Θ] - [1/sin^2Θ])
2([1 - 2cos^2Θ]/cos^2Θ) / ([sin^2Θ - cos^2Θ]/cos^2Θsin^2Θ)
-2(2cos^2Θ -1) / -1([cos^2Θ - sin^2Θ] / sin^2Θ)
* note : 2cos^2Θ -1 = cos2Θ and cos^2Θ - sin^2Θ = cos2Θ
*note : minus signs get cancelled
2cos2Θ / (cos2Θ/sin^2Θ)
2/(1/sin^2Θ)
2sin^2Θ
=RHS.
just realised this z a longer way to do it..the 1st answer is pretty simple. :)
(tanΘ-cotΘ)/(tanΘ+cotΘ) + 1 = 2sin²Θ?
L.H.S. = (tanΘ-cotΘ)/(tanΘ+cotΘ) + 1 =
= (tanΘ-1/tanΘ) / (tanΘ+1/tanΘ) +1 = (tan^2Θ-1) / (tan^2Θ+1)+1 =
= (tan^2Θ-1 + tan^2Θ+1) / (tan^2+1) = 2tan^2Θ / sec^2Θ =
= (2sin²Θ/cos^2Θ) / (1/cos^2Θ) =
= 2sin²Θ = R.H.S. >========================< Q . E . D
1) By relation we have cot(θ) = 1/{tan(θ)}
So multiplying Nr. & Dr. of the 1st term by tan(θ),
left side = [(tan²θ - 1)/(tan²θ + 1)] + 1
= [(tan²θ - 1)/(sec²θ)] + 1; since tan²θ + 1 = sec²θ.
= [(tan²θ/sec²θ) - (1/sec²θ)] + 1
= (sin²θ - cos²θ) + 1 [since, sec²θ = 1/cos²θ and tan²θ = sin²θ/cos²θ]
= 2sin²θ - 1 + 1 [since cos²θ = (1 - sin²θ)]
= 2sin²θ = Right side [Proved]
(tan(t) - cot(t)) / (tan(t) + cot(t)) => (sin(t)/cos(t) - cos(t)/sin(t)) / (sin(t)/cos(t) + cos(t)/sin(t)) => ((sin(t)^2 - cos(t)^2) / (sin(t)cos(t))) / ((sin(t)^2 + cos(t)^2) / (sin(t)cos(t))) => (sin(t)^2 - cos(t)^2) / (sin(t)^2 + cos(t)^2) Now, sin(t)^2 + cos(t)^2 = a million for all values of t. Now we are left with: (sin(t)^2 - cos(t)^2) / a million => sin(t)^2 - cos(t)^2
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
Unless you can see any nice method to do this, the best way to go is to convert everything in terms of sines and cosines.
Using tanθ = sinθ/cosθ and cotθ = sinθ/cosθ, the left side becomes:
LHS = (tanθ - cotθ)/(tanθ + cotθ) + 1
= (sinθ/cosθ - cosθ/sinθ)/(sinθ/cosθ + cosθ/sinθ) + 1
= (sin^2θ - cos^2θ)/(sin^2θ + cos^2θ) + 1, by multiplying top/bottom by sinθcosθ
= sin^2θ - cos^2θ + 1, since sin^2θ + cos^2θ = 1
= sin^2θ + (1 - cos^2θ), by re-arranging
= sin^2θ + sin^2θ, since sin^2θ + cos^2θ = 1
= 2sin^2θ
= RHS.
I hope this helps!
(tanΘ-cotΘ)/(tanΘ+cotΘ) + 1 = 2sin²Θ
LHS =
{(tanΘ-cotΘ)/(tanΘ+cotΘ)} +1
{(tanΘ-cotΘ)(tanΘ-cotΘ)/(tanΘ+cotΘ)(tanΘ-cotΘ)} +1
{(tanΘ-cotΘ)^2 / (tan^2Θ - cot^2Θ)} +1
{(tan^2Θ + cot^2Θ - 2tanΘcotΘ) / (tan^2Θ - cot^2Θ)} +1
* note : tanΘ.cotΘ = 1
{(tan^2Θ + cot^2Θ - 2) / (tan^2Θ - cot^2Θ)} +1
*note : taking (tan^2Θ - cot^2Θ) as the common denominator
(tan^2Θ + cot^2Θ - 2 + tan^2Θ - cot^2Θ) / (tan^2Θ - cot^2Θ)
(2tan^2Θ - 2) / (tan^2Θ - cot^2Θ)
*note : tan^2Θ +1 = sec^2Θ and cot^2Θ +1 = cosec^2Θ
(2sec^2Θ - 4) / (sec^2Θ - cosec^2Θ)
*note : sec^2Θ = 1/cos^2Θ and cosec^2Θ= 1/sin^2Θ
2([1/cos^2Θ] - 2) / ([1/cos^2Θ] - [1/sin^2Θ])
2([1 - 2cos^2Θ]/cos^2Θ) / ([sin^2Θ - cos^2Θ]/cos^2Θsin^2Θ)
-2(2cos^2Θ -1) / -1([cos^2Θ - sin^2Θ] / sin^2Θ)
* note : 2cos^2Θ -1 = cos2Θ and cos^2Θ - sin^2Θ = cos2Θ
*note : minus signs get cancelled
2cos2Θ / (cos2Θ/sin^2Θ)
2/(1/sin^2Θ)
2sin^2Θ
=RHS.
just realised this z a longer way to do it..the 1st answer is pretty simple. :)
(tanΘ-cotΘ)/(tanΘ+cotΘ) + 1 = 2sin²Θ?
L.H.S. = (tanΘ-cotΘ)/(tanΘ+cotΘ) + 1 =
= (tanΘ-1/tanΘ) / (tanΘ+1/tanΘ) +1 = (tan^2Θ-1) / (tan^2Θ+1)+1 =
= (tan^2Θ-1 + tan^2Θ+1) / (tan^2+1) = 2tan^2Θ / sec^2Θ =
= (2sin²Θ/cos^2Θ) / (1/cos^2Θ) =
= 2sin²Θ = R.H.S. >========================< Q . E . D
1) By relation we have cot(θ) = 1/{tan(θ)}
So multiplying Nr. & Dr. of the 1st term by tan(θ),
left side = [(tan²θ - 1)/(tan²θ + 1)] + 1
= [(tan²θ - 1)/(sec²θ)] + 1; since tan²θ + 1 = sec²θ.
= [(tan²θ/sec²θ) - (1/sec²θ)] + 1
= (sin²θ - cos²θ) + 1 [since, sec²θ = 1/cos²θ and tan²θ = sin²θ/cos²θ]
= 2sin²θ - 1 + 1 [since cos²θ = (1 - sin²θ)]
= 2sin²θ = Right side [Proved]
(tan(t) - cot(t)) / (tan(t) + cot(t)) => (sin(t)/cos(t) - cos(t)/sin(t)) / (sin(t)/cos(t) + cos(t)/sin(t)) => ((sin(t)^2 - cos(t)^2) / (sin(t)cos(t))) / ((sin(t)^2 + cos(t)^2) / (sin(t)cos(t))) => (sin(t)^2 - cos(t)^2) / (sin(t)^2 + cos(t)^2) Now, sin(t)^2 + cos(t)^2 = a million for all values of t. Now we are left with: (sin(t)^2 - cos(t)^2) / a million => sin(t)^2 - cos(t)^2