Math, asked by asadali1407p44y4s, 11 months ago

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Answered by mamatha83
0
Taking LHS,we have 
=cot²A(secA-1)/1+sinA 
=cot²A(secA-1)(1-sinA)/((1+sinA)(1-sinA...‡ 
=cot²A(secA-1)(1-sinA)/(cos²A) 
=(secA-1)(1-sinA)/(sin²A) 
=(secA-1)(secA+1)(1-sinA)/(sin²A)(secA+...‡ 
=(sec²A-1)(1-sinA)/(sin²A)(secA+1) 
=(tan²A)(1-sinA)/(sin²A)(secA+1) 
=((sin²A)/(cos²A)(1-sinA))/(sin²A)(secA...‡ 
=(1-sinA)/((secA+1)(cos²A)) 
=sec²A(1-sinA)/(1+secA)=RHS 
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