Advertisements
Advertisements
प्रश्न
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
Advertisements
उत्तर
LHS = `(1- tan^2 theta)/(1+ tan^2 theta)`
=`(1-(sin^2 theta)/(cos^2 theta))/(1+(sin^2 theta)/(cos^2 theta))`
=`(cos^2 theta- sin^2 theta)/(cos^2 theta+ sin ^2 theta)`
=`(cos^2 theta+sin^2 theta)/1`
=`cos^2 theta- sin^2 theta`
= RHS
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
` tan^2 theta - 1/( cos^2 theta )=-1`
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
If 5 `tan theta = 4,"write the value of" ((cos theta - sintheta))/(( cos theta + sin theta))`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Evaluate:
`(tan 65°)/(cot 25°)`
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
sin(45° + θ) – cos(45° – θ) is equal to ______.
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
