Question

cot^2theta (sectheta - 1)/ 1 + sintheta + sec^2 theta ( sin theta - 1) / 1 + sec theta = 0​

Answer 1

Solution:

Q: cot²θ (sec θ - 1)/(1 + sin θ) + sec²θ (sin θ - 1)/(1 + sec θ) = 0

Considering L.H.S:

= cot²θ (sec θ - 1)/(1 + sin θ) + sec²θ (sin θ - 1)/(1 + sec θ)

• Taking minus common,

= cot²θ (sec θ - 1)/(1 + sin θ) - sec²θ (1 - sin θ)/(sec θ + 1)

= cot²θ (sec θ - 1)(sec θ + 1)/(1 + sin θ)(sec θ + 1) - sec²θ (1 - sin θ)(sec θ - 1)/(sec θ + 1)(sec θ - 1)

= cot²θ (sec²θ - 1)/(1 + sin θ)(sec θ + 1) - sec²θ (1 - sin θ)(sec θ - 1)/(sec²θ - 1)

• We know: sec²θ - 1 = tan²θ

= cot²θ • tan²θ/(1 + sin θ)(sec θ + 1) - sec²θ (1 - sin θ)(sec θ - 1)/tan²θ

• We know: cot²θ • tan²θ = 1
• sec²θ/tan²θ = cosec²θ

= 1/(1 + sin θ)(sec θ + 1) - cosec²θ(1 - sin θ)(sec θ - 1)/1

• Taking (1 + sin θ)(sec θ + 1) as L.C.M,

= {1 - cosec²θ(1 - sin θ)(sec θ - 1)•(1 + sin θ)(sec θ + 1)}/(1 + sin θ)(sec θ + 1)

= {1 - cosec²θ(1 - sin²θ)(sec²θ - 1)}/(1 + sin θ)(sec θ + 1)

• We know: 1 - sin²θ = cos²θ
• sec²θ - 1 = tan²θ

= {1 - cosec²θ•cos²θ•tan²θ}/(1 + sin θ)(sec θ + 1)

• Here: cosec²θ•cos²θ•tan²θ = 1

= {1 - 1}/(1 + sin θ)(sec θ + 1)

= 0/(1 + sin θ)(sec θ + 1)

= 0

= R.H.S

Hence, Proved.

Answer 2

Prove that :-

$\tt \: \dfrac{ {cot}^{2} \theta \: (sec\theta \: - 1) }{1 + sin\theta \: } + \dfrac{ {sec}^{2}\theta \: (sin\theta \: - 1) }{sec\theta \: + 1}$

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

$1. \: \boxed{ \pink{ \rm \:1 - {cos}^{2}\theta \: = {sin}^{2} \theta \: }}$

$2. \: \boxed{ \pink{ \rm \: {sec}^{2} \theta \: - 1 = {tan}^{2}\theta \: }}$

$3. \: \boxed{ \pink{ \rm \:cot\theta \: \times tan\theta \: = 1 }}$

$4. \: \boxed{ \pink{ \rm \: \dfrac{1}{cos\theta \: } \: = \: sec\theta \: }}$

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

Consider

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

On rationalizing the numerator and denominator, we get

$\rm : \implies \: \dfrac{ {cot}^{2} \theta \: (sec\theta \: - 1) }{1 + sin\theta \: } \times \dfrac{sec\theta \: + 1}{1 - sin\theta \: } \times \dfrac{1 - sin\theta \: }{sec\theta \: + 1}$

$\rm : \implies \: \dfrac{ {cot}^{2} \theta \: ( {sec}^{2} \theta \: - 1) }{1 - {sin}^{2} \theta \: } \times \dfrac{1 - sin\theta \: }{sec\theta \: + 1}$

$\rm : \implies \: \dfrac{ {cot}^{2} \theta \: \times {tan}^{2} \theta \: }{ {cos}^{2}\theta \: } \times \dfrac{1 - sin\theta \: }{sec\theta \: + 1}$

$\rm : \implies \: \dfrac{1}{ {cos}^{2} \theta \: } \times \dfrac{1 - sin\theta \: }{sec\theta \: + 1}$

$\tt : \implies \: {sec}^{2} \theta \: \times \dfrac{1 - sin\theta \: }{sec\theta \: + 1}$

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

$\boxed{ \pink{ \bf \: Hence, \: \dfrac{ {cot}^{2} \theta \: (sec\theta \: - 1) }{1 + sin\theta \: } = \: - \: \dfrac{ {sec}^{2}\theta \: (sin\theta \: - 1) }{sec\theta \: + 1} }}$

On transposition, we get

$\rm : \implies \: \dfrac{ {cot}^{2} \theta \: (sec\theta \: - 1) }{1 + sin\theta \: } \: + \: \dfrac{ {sec}^{2}\theta \: (sin\theta \: - 1) }{sec\theta \: + 1} = 0$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

$\large \red{\bf \: ⟼ Explore \: \: more }$

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