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प्रश्न
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
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उत्तर
LHS = `sqrt(2 + tan^2 θ + cot^2 θ)`
= `sqrt( tan^2 θ + cot^2θ + 2tan θ.cot θ)` ...[ ∵ tan θ.cot θ = 1 ]
= `sqrt( tan^2 θ + cot^2θ)`
= tan θ + cot θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`(1 + cos theta + sin theta)/(1 + cos theta - sin theta) = (1 + sin theta)/cos theta`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Choose the correct alternative:
cos θ. sec θ = ?
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
