If tanA +sinA= m and tanA-sinA=n, then prove that (m+n)²= {16nm/(m-n)²}
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Answer:
Step-by-step explanation:
tanA+sinA= m ---(1)
tanA-sinA = n ---(2)
i) m+n = 2tanA ---(3)
ii)m-n=2sinA ----(4)
iii) m²-n²
=(m+n)(m-n) /* From (3)&(4)*/
= 2tanA × 2sinA
= 4tanAsinA ---(5)
Now,
LHS = (m²-n²)²
= (4tanAsinA)²
= 16tan²Asin²A
= 16tan²A(1-cos²A)
=16[tan^{2}-tan^{2}Acos^{2}A}]
=16[tan^{2}A-\frac{sin^{2}A cos^{2}A}{cos^{2}A}]
=16[tan^{2}A-sin^{2}A}]
=16(tanA+sinA)(tanA-sinA)
=16mn\:[from\:(1)\:and\:(2) ]
=RHS
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