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рдкреНрд░рд╢реНрди
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
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рдЙрддреНрддрд░
LHS= `(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))`
=`(cos^2 theta/sin^2 theta(1/costheta-1))/((+ sin theta)) + (1/cos^2 theta(sin theta -1))/((1+ 1/cos theta))`
=`((cos^2 theta)/(sin^2 theta )((1- cos theta)/(cos theta)))/((1+sin theta))+ (((sin theta -1 ))/(cos ^2theta ))/(((cos theta + 1 )/(cos theta)))`
=`(cos^2 theta (1- cos theta))/(sin^2 theta cos theta (1+ sin theta))+ ((sin theta -1) cos theta)/((cos theta +1 ) cos^2 theta)`
=`(cos theta (1-cos theta))/((1- cos^2 theta)(1+ sin theta)) + ((sin theta -1)cos theta)/((costheta + 1 ) (1- sin^2 theta))`
=`(cos theta (1-cos theta))/((1- cos theta )( 1+ cos theta )(1+ sin theta)) + (-(1 sin theta ) cos theta)/((cos theta +1)(1-sin theta )(1+ sin theta))`
=`cos theta/((1+ cos theta )(1+ sin theta)) - cos theta/((cos theta +1)(1+ sin theta))`
= ЁЭЬГ
= RHS
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sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
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Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
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`1/(tan A + cot A) = cos A sin A`
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(sec A − tan A)2 (1 + sin A) = (1 − sin A)
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Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
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