Question

Evaluate : (1+tanA+ secA) (1+cotA-cosecA)
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Answer 1

GIVEN :

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

We know that,

tanA = sinA/cosA

secA = 1/cosA

cotA = cosA/sinA

cosecA = 1/sinA

= (1 + sinA/cosA + 1/cosA) (1 + cosA/sinA - 1/sinA)

= (1 + sinA + 1/ cosA ) (1 + cosA - 1 / sinA)

After LCM,

= (cosA + sinA + 1)/cosA (sinA + cosA - 1)/sinAcosA

= (cosA × cosA) + (sinA × sinA) + 1 × (- 1 ) / sinAcosA

= (cosA + sinA)² - 1 / sinAcosA

= cos²A + sin²A + 2SinAcosA - 1 / sinAcosA

We know that,

cos²A + sin²A = 1

= 1 - 1 + 2sinAcosA / sinAcosA

= 2sinAcosA/sinAcosA

= 2

Therefore,

Therefore,(1+tanA+ secA)(1+cotA-cosecA)= 2

Answer 2

Answer:

(1+tanA+ secA) (1+cotA-cosecA) = 2

Explanation:

Given Problem:

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

Solution:

$\textbf(1+\dfrac{sinA}{cosA} + \dfrac{1}{CosA}\textbf)\: \textbf(1+\dfrac{cosA}{sinA} -\dfrac{1}{SinA}\textbf)$

$=\textbf(\dfrac{CosA+SinA+1}{CosA}\textbf )\ \textbf (\dfrac{SinA+CosA-1}{SinA}\textbf)$

$\textbf =\textbf[\dfrac{\textbf(cosA+SinA\textbf)^2-1}{CosA\:SinA}\textbf]$

$\textbf=\textbf\ [\dfrac{cos^2A+sin^2A+2cosASinA-1}{cosAsinA}\textbf]$

$\textbf=\textbf[\dfrac{1+2cosAsinA-1}{cosASinA}\textbf]$

$\textbf=\textbe[ \dfrac{2cosASinA}{cosASinA}]$

= 2

Hence,

(1+tanA+ secA) (1+cotA-cosecA) = 2