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question..
( 1+tan2A/1+cos²A) =(1-tanA/1-cotA)² =tan²A
answer.....
★Given that: 1+tan²A1+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★
★hope.it.helps....★
thanks....★
Answer:
1 = tan is answer please make me brainslit please