Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Advertisements
उत्तर
`(cotA + cosecA - 1)/(cotA - cosecA + 1)`
= `(cotA + cosecA - (cosec^2A - cot^2A))/(cotA - cosecA + 1)` [`cosec^2A - cot^2A = 1`]
= `(cotA + cosecA - [(cosecA - cotA)(cosecA + cotA)])/(cotA - cosecA + 1)`
= `(cotA + cosecA[1 - cosecA + cotA])/(cotA - cosecA + 1)`
= `cotA + cosecA`
= `cosA/sinA + 1/sinA`
= `(1 + cosA)/sinA`
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Prove that `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`.
Prove that `(cot A + "cosec" A - 1)/(cot A - "cosec" A + 1) = (1 + cos A)/(sin A)`.
