Advertisements
Advertisements
प्रश्न
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
उत्तर
LHS = sec2 (90° - θ) + tan2 (90° - θ)
= cosec2θ + cot2θ
= 1 + cot2θ + cot2θ
= 1 + 2cot2θ
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
Without using trigonometric table , evaluate :
`cosec49°cos41° + (tan31°)/(cot59°)`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`