Question

(1+secA+tanA)(1+cotA-cosecA)​

Answer 1

Step-by-step explanation:

Solution: Given: (1+ tan A + sec A) (1+ cot A - cosec A)

\\=\left (1+\frac{\sin A}{\cos A}+\frac{1}{\cos A} \right )\left (1+\frac{\cos A}{\sin A}-\frac{1}{\sin A} \right )\\\\\\ =\left (\frac{\cos A+\sin A+1}{\cos A} \right )\left (\frac{\sin A+\cos A-1}{\sin A} \right )\\\\\\ =\frac{(\sin A+\cos A)^2-1^2}{\sin A.\cos A} \quad\quad [\because (a-b)(a+b)=a^2-b^2]\\\\\\ =\frac{\sin^2 A+\cos^2 A+2\sin A.\cos A-1}{\sin A.\cos A} \quad\quad [\because (a+b)^2=a^2+b^2+2ab]\\\\\\ =\frac{1+2\sin A.\cos A-1}{\sin A.\cos A} \quad\quad [\because \sin^2 A+\cos^2 A=1]\\\\\\ =\frac{2\sin A.\cos A}{\sin A.\cos A}\\\\ =2

Answer 2

Step-by-step explanation:

ur answer is in the attachment

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