Answer :



Proof:


Identities used:



Therefore,


tan 15° = tan (45° - 30°)







On rationalising:




{ (a – b)(a + b) = a2 – b2}








On rationalising




{ (a – b)(a + b) = a2 – b2}




Let 2θ = 15°



We know,






Formula used:





{ (a + b)2 = a2 + b2 + 2ab}




cot θ < 0 as θ is in 1st quadrant.


So,




{ (a + b)2 = a2 + b2 + 2ab}








{ cot θ = tan(90° - θ)}



Hence Proved


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