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प्रश्न
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
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उत्तर
`(tan^2 theta - sec^2 theta )/ (cot^2 theta - cosec^2 theta)`
=` (-1)/(-1)`
= 1
APPEARS IN
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
(i)` (1-cos^2 theta )cosec^2theta = 1`
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
`sin theta/((cot theta + cosec theta)) - sin theta /( (cot theta - cosec theta)) =2`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
The value of sin2θ + `1/(1 + tan^2 theta)` is equal to
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`
If tan θ = `x/y`, then cos θ is equal to ______.
