Question

(1+tanA+ secA) (1+ cotA-cosecA) =​

Answer 1

hey mate here is your answer

Answer 2

Given:- (1+tanA+ secA) (1+ cotA-cosecA)

To Find:- The value of (1+tanA+ secA) (1+ cotA-cosecA)

Solution:-

$=(1+\frac{sin A}{cos A}+\frac{1}{cos A})(1+\frac{cos A}{sin A}- \frac{1}{sin A})\\\\=(\frac{cos A+ sin A+1}{cos A})(\frac{sin A+cos A-1}{sin A})\\\\=\frac{ (sin A+cos A)^{2} -1^{2} }{sinA*cos A} \ [since,(a-b)(a+b)=a^{2}-b^{2}]\\ \\ =\frac{sin^{2} A+cos^{2}A+2sinA*cosA-1 }{sinA*cosA} \ [since,(a+b)^{2}=a^{2}+b^{2}+2ab]\\\\=\frac{1+2sinA*cosA-1}{sinA*cosA} \ [since,sin^{2}A+cos^{2}A=1]\\\\=\frac{2sinA*cosA}{sinA*cosA} \\\\=2$

Therefore, the answer is 2.

Answer:- The value of (1+tanA+ secA) (1+ cotA-cosecA) is 2.