Advertisements
Advertisements
प्रश्न
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
Advertisements
उत्तर
To Prove: `((1 + tan^2 A))/((1 + cot^2 A))` = sec2 A – 1
LHS.
We have, `(((1 + sin^2 A)/(cos^2 A)))/(((1 + cos^2 A)/(sin^2 A)))`
= `[(((cos^2 A + sin^2 A))/(cos^2 A))/(((sin^2 A + cos^2 A))/(sin^2 A))]`
= `((1/cos^2 A))/((1/sin^2 A))` ...[As sin2 A + cos2 A = 1]
= `((sin^2 A))/((cos^2 A))`
= tan2 A
= sec2 A – 1
Hence, proved.
APPEARS IN
संबंधित प्रश्न
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
Write the value of tan10° tan 20° tan 70° tan 80° .
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Prove that `(tan^2"A")/(tan^2 "A"-1) + (cosec^2"A")/(sec^2"A"-cosec^2"A") = (1)/(1-2 co^2 "A")`
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
