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Prove that: cosec theta/ 1+sec theta + 1 sec theta/cosec theta = 2cosec^3 theta [ sec theta 1]?
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Prove that: cosec theta/ 1+sec theta + 1 sec theta/cosec theta = 2co...
Proof: cosec theta/ 1 sec theta 1 sec theta/cosec theta = 2cosec^3 theta [ sec theta 1]


Step 1: Rewrite the expression

Start by writing the left-hand side of the equation in terms of sines and cosines:
cosec theta/ 1 sec theta 1 sec theta/cosec theta
= (1/sin theta) / (1/cos theta) (1/cos theta) / (1/sin theta)
= cos theta / sin theta sin theta / cos theta

Step 2: Simplify

Next, simplify each fraction:
cos theta / sin theta sin theta / cos theta
= (cos theta)^2 / (sin theta)(cos theta) (sin theta)^2 / (sin theta)(cos theta)
= cos theta / sin^2 theta 1 / sin theta

Step 3: Combine

Combine the two fractions:
cos theta / sin^2 theta 1 / sin theta
= cos theta / sin^2 theta + sin theta / sin^2 theta
= (cos theta + sin theta) / sin^2 theta

Step 4: Use trigonometric identities

Use the trigonometric identity (cos x + sin x)^2 = cos^2 x + 2cos x sin x + sin^2 x to simplify the numerator:
(cos theta + sin theta)^2 = cos^2 theta + 2cos theta sin theta + sin^2 theta
= 1 + 2cos theta sin theta

Step 5: Final simplification

Substitute this back into the expression and simplify:
(cosec theta/ 1 sec theta) (1 sec theta/cosec theta)
= (cos theta + sin theta)^2 / (sin^2 theta)
= (1 + 2cos theta sin theta) / (sin^2 theta)
= 2(cos theta / sin theta)^3
= 2cosec^3 theta

Therefore, cosec theta/ 1 sec theta 1 sec theta/cosec theta = 2cosec^3 theta [ sec theta 1].
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Prove that: cosec theta/ 1+sec theta + 1 sec theta/cosec theta = 2co...
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Prove that: cosec theta/ 1+sec theta + 1 sec theta/cosec theta = 2cosec^3 theta [ sec theta 1]?
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