Question

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Answer 1

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Answer 2

(cosA - sinA + 1)÷(cosA + sinA - 1 )= cosecA + cot A

LHS = (cosA - sinA + 1) ÷ ( cosA + sinA - 1)

(Divide numerator and denominator by sinA)

= [(cosA-sinA+1)÷sinA]÷[(cosA+sinA-1)÷sinA]

= [(cosA÷sinA)-(sinA÷sinA)+(1÷sinA)] ÷ [(cosA÷sinA)+(sinA÷sinA)-(1÷sinA)]

((cosA÷sinA) = cotA, (1÷sinA)= cosecA)

= (cotA-1+cosecA)÷(cotA+1-cosecA)

(Use trigonometric identity cosec²A-cot²A=1)

= [cotA-(cosec²A-cot²A)+cosecA]÷(cotA+1-cosecA)

= [(cotA+cosecA)-(cosec²A-cot²A)]÷(cotA+1-cosecA)

(Use algebraic identity a²+b² = (a+b)(a-b))

= [(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)]÷(cotA+1-cosecA)

(Take cotA+cosecA common)

=[(cotA+cosecA)(1-(cosecA-cotA))]÷(cotA+1-cosecA)

= [(cotA+cosecA)(cotA+1-cosecA)]÷(cotA+1-cosecA)

= cotA+cosecA

Hence proved.

☆Remember that trigonometry is a kind of subject which needs a lot of practise to get solved.It cannot be solved without remembering the steps !

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