Question

tanθ/secθ-1 + tanθ/secθ+1 =​

Answer 1

Hi mate!!

Hi mate!!Go through the above attachment!!

Answer 2

$\begin{gathered}\Large{\bold{\blue{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \tt \: {sec}^{2} \theta \: - {tan}^{2} \theta \: = 1$

$(2). \: \tt \: sec\theta \: = \dfrac{1}{cos\theta \:}$

$(3). \: \tt \: tan\theta \: = \dfrac{sin\theta \:}{cos\theta \:}$

$(4). \: \tt \: cosec\theta \: = \dfrac{1}{sin\theta \:}$

$\large\underline\purple{\bold{Solution :- }}$

$\rm :\implies\:\dfrac{tan \theta \: }{sec\theta \: - 1} + \dfrac{tan\theta \:}{sec\theta \: + 1}$

On taking LCM, we get

$\rm :\implies\:\dfrac{tan\theta \:(sec\theta \: + 1) + tan\theta \:sec\theta \: - 1)}{(sec\theta \: + 1) (sec\theta \: - 1) }$

$\rm :\implies\:\dfrac{tan\theta \:(sec\theta \: + \cancel1 + sec\theta \: - \cancel1)}{ {sec}^{2}\theta \: - 1 }$

$\rm :\implies\:\dfrac{2sec\theta \:tan\theta \:}{ {tan}^{2}\theta \: }$

$\rm :\implies\:\dfrac{2sec\theta \:}{tan\theta \:}$

$\rm :\implies\:2 \times \dfrac{1}{cos\theta \:} \times \dfrac{cos\theta \:}{sin\theta \:}$

$\rm :\implies\:\dfrac{2}{sin\theta \:}$

$\rm :\implies\:2 \: cosec\theta \:$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1