Question

( 1+ cotA – cosecA)(1 + tanA + secA) = 2​

Answer 1

$\maltese$Given to prove that:-

$(1+cotA-cscA)(1+tanA+secA) = 2$

$\maltese$Solution :-

As we know that,

$cotA = \dfrac{cosA}{sinA}$

$cscA = \dfrac{1}{sinA}$

$tanA = \dfrac{sinA}{cosA}$

$secA =\dfrac{1}{cosA}$

$\maltese$Substituting the values ,

$\bigg(1+\dfrac{cosA}{sinA} - \dfrac{1}{sinA} \bigg)\bigg(1+\dfrac{sinA}{cosA} +\dfrac{1}{cosA} \bigg)$

$\bigg(1+\dfrac{cosA-1}{sinA}\bigg)\bigg(1+\dfrac{sinA+1}{cosA} \bigg)$

$\maltese$Take L.C.M to the denominator

$\bigg(\dfrac{sinA+cosA-1}{sinA}\bigg)\bigg(\dfrac{cosA+sinA+1}{cosA} \bigg)$

$\dfrac{(sinA+cosA+1)(sinA+cosA-1)}{(sinA)(cosA)}$

$\maltese$Simplifying the numerator by applying formula (a-b)(a+b)= a^2-b^2

$\dfrac{(sinA+cosA)^2-1}{sinAcosA}$

$\dfrac{sin^2A+cos^2A+2sinAcosA-1}{sinAcosA}$

$\dfrac{1+2sinAcosA-1}{sinAcosA}$

$\dfrac{\not1+2sinAcosA\not{-1}}{sinAcosA}$

$\dfrac{2sinAcosA}{sinAcosA}$

$=2$

$\maltese$Hence proved!

$\maltese$Used formulae:-

• sin²A + cos² A = 1
• cotA = cosA/sinA
• cscA = 1/sinA
• tanA = sinA/cosA
• secA = 1/cosA

$\maltese$ Know more formulae :-

Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj