Advertisements
Advertisements
प्रश्न
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
Advertisements
उत्तर
LHS = `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ))`
LHS = `(sin^2 θ/cos θ). (cos^2 θ/sin θ)`
LHS = sin θ. cos θ
RHS = `1/(tan θ + cot θ)`
RHS = `1/((sin^2 θ + cos^2 θ)/(sin θ. cos θ))`
RHS = `(sin θ. cos θ)/(sin^2 θ + cos^2 θ)`
RHS = sin θ. cos θ
LHS = RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
(sec A + tan A) (1 − sin A) = ______.
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
