show that 1-tan² A / cot²A-1= tan²A
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Given that: 1+tan² A1+cot²A=[1-tanA1-cotA]²=tan²A
We will first solve the equation on LHS
LHS:
= 1+tan²A / 1+cot²A
Using the trignometric identities we know that 1+tan²A= Sec²A and 1+cot²A= Cosec²A
= Sec²A/ Cosec²A
On taking the reciprocals we get
= Sin²A/Cos²A
= tan²A
RHS:
=(1-tanA)²/(1-cotA)²
Substituting the reciprocal value of tan A and cot A we get,
=(1-sinA/cosA)²/(1-cosA/sinA)²
=[(cosA-sinA)/cosA]²/ [(sinA-cos)/sinA)²
=(cosA-sinA)²×sin²A /Cos²A. /(sinA-cosA)²
=1×sin²A/Cos²A×1.
=tan
The values of LHS and RHS are same.
Hence proved
Step-by-step explanation:
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Step-by-step explanation:
As we know that SinA/CosA= TanA
Taking .Laft hand side
in second step
=1-Sin^2A/Cos^2A/Cos^2A/Sin^2A-1
= Cos^2A-sin^2A/Cos^2A/ Cos^2A-Sin^2A/ Sin^2A
the nominator tems will be cancal with each other
=Sin^2A/Cos^2A= Tan^2A proof
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