Prove the following identities, where the angles involved are acute angles for which the expressions are defined.


(i)


(ii)


(iii)


(iv)


(v) using the identity


(vi)


(vii)


(viii)


(ix) [Hint: Simplify LHS and RHS separately]


(x)


(i)


 


LHS = (cosec θ - cot θ)


 


= ()2


 


=


 


= (1 - cos θ)2/sin2 θ


 


= (1 – cos θ)2/1 – cos2 θ


 


= (1 – c0s θ)2/ (1 – cos θ) (1 + cos θ)


 


=


 


= RHS


 


(ii)


 


LHS = +


 


= cos2A + (1 + sin2A)/(1 + sinA) (cos A)


 


= cos2A + 1 + sin2A + 2sinA/(1 + sinA) (cosA)


 


= sin2A + cos2A + 1 + 2sinA/(1 + sinA) (cosA)


 


=


 


=


 


=


 


=


 


= 2 sec A


 


= RHS


 


(iii)


 


[Hint: Write the expression in terms of sin θ and Cos θ]


 



 


=


 


=


 


= sin2 θ/[cos θ (sin θ – cos θ)] + cos2 θ/ [sin θ (cos θ – sin θ)]


 


= sin2 θ/[cos θ (sin θ - cos θ)] – cos2 θ/[sin θ (sin θ - cos θ)]


 


= 1/(sin θ - cos θ) [(sin2 θ/cos θ) – (cos2 θ/sin θ)]


 


= 1/(sin θ – cos θ) * [(sin 3 θ – cos3 θ)/sin θ cos θ]


 


= [(sin θ – cos θ) (sin2 θ + cos2 θ + sin θcos θ)]/ [sin θ - cos θ) sin θ cos θ]


 


= (1 + sin θ cos θ)/sin θ cos θ


 


= 1/sin θ cos θ + 1


 


= 1 + sec θ cosec θ


 


= RHS


 


(iv)


 


[Hint: Simplify LHS and RHS separately]


 


LHS = (1 + secA)/ secA


 


= (1 + 1/cos A)/1/cos A


 


= (cos A + 1)/cosA/1/cosA


 


= cos A + 1


 


RHS = sin2A/(1 – cos A)


 


= (1 – cos2A)/(1 – cos A)


 


= (1 – cos A) (1 + cos A)/ (1 – cos A)


 


= cos A + 1


 


LHS = RHS


 


(v) using the identity


 


LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)


 


Dividing Numerator and Denominator by sin A


 


= (c0s A – sin A + 1) / sin A/ (cos A sin A – 1) / sin A


 


= (cot A – 1 + cosec A)/ (cot A + 1 – cosec A)


 


= (cot A – cosec2A + cot2A + cosec A)/ (cot A + 1- cosec A) [Using cosec2A – cot2A = 1]


 


= [(cot A + cosec A) – (cosec A + cot A) (cosec A – cot A)]/ (1 – cosec A + cot A)


 


= (cot A + cosec A) (1 – cosec A + cot A)/ (1 – cosec A + cot A)


 


= cot A + cosec A


 


= RHS


 


(vi)


 


Dividing numerator and denominator of LHS by cos A


 


=


 


=


 


= *


 


= [(sec A + tan A)2/sec2A – tan2A]1/2


 


=


 


= sec A + tan A


 


= RHS


 


(vii)


 


(sin θ – 2 sin3 θ)/ (2cos3 θ – cos θ)


 


= [sin θ (1 – 2sin2 θ)]/[cos θ (2cos2 θ – 1)]


 


= sin θ [1 – 2 (1 – cos2 θ)] / [cos θ (2 cos2 θ – 1)]


 


= [sin θ (2cos2 θ – 1)] / [cos θ (2cos2 θ – 1)]


 


= tan θ = RHS


 


(viii)


 


LHS = (sin A + cosec A)2 + (cos A + sec A)2


 


= (sin2A + cosec2A + 2sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)


 


= (sin2A + cos2A) + 2 sin A (1/sinA) + 2cos A (1/cosA) + 1 + tan2A + 1 + cot2A


 


= 1 + 2 + 2 + 2 + tan2A + cot2A


 


= 7 + tan2A + cot2A


 


= RHS


 


(ix)


 


[Hint: Simplify LHS and RHS separately]


 


LHS = (cosec A – sin A) (sec A – cos A)


 


= (1/sin A – sin A) (1/cos A – cos A)


 


= [(1 – sin2A)/ sinA] [(1 – cos2A)/cosA]


 


= (cos2A/sin A) * (sin2A/cos A)


 


= cos A sin A


 


RHS = 1/(tan A + cot A)


 


= 1 / (sin A/cos A + cos A / sin A)


 


= 1/[(sin2A + cos2A)/sin A cos A]


 


= cos A sin A


 


LHS = RHS


 


(x)


 


LHS = (1 + tan2A/ 1 + cot2A)


 


= (1 + tan2A / 1 + 1/tan2A)


 


= 1 + tan2A/ [(1 + tan2A)/tan2A]


 


= tan2A


 

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