Question

Prove the following trigonometric ratios: (its urgent)

Answer 1

EXPLANATION.

(1) = Cos²∅ - Cos²∅Sin²∅ = Cos⁴∅.

As we know that,

⇒ Sin²∅ = 1 - Cos²∅.

Put the Formula in equation, we get.

⇒ Cos²∅ - Cos²∅(1 - Cos²∅).

⇒ Cos²∅ - Cos²∅ + Cos⁴∅.

⇒ Cos⁴∅.

Hence L.H.S = R.H.S.

(2) = (1 - Cos²∅)(1 + Tan²∅) = Tan²∅.

As we know that,

⇒ 1 - Cos²∅ = Sin²∅.

⇒ (Sin²∅)(1 + Tan²∅).

⇒ Sin²∅ + Sin²∅Tan²∅.

⇒ Sin²∅ + Sin²∅.Sin²∅/Cos²∅.

⇒ Sin²∅ + Sin⁴∅/Cos²∅.

Taking L.C.M in the equation, we get.

⇒ Sin²∅.Cos²∅ + Sin⁴∅/Cos²∅.

⇒ Sin²∅(Cos²∅ + Sin²∅)/Cos²∅.

As we know that,

⇒ Cos²∅ + Sin²∅ = 1.

⇒ Sin²∅/Cos²∅.

⇒ Tan²∅.

Hence, L.H.S = R.H.S.

(3) = (1 + Cot²A)(1 - Sin²A) = Cot²A.

As we know that,

⇒ 1 - Sin²∅ = Cos²∅.

⇒ (1 + Cot²A)(Cos²A).

⇒ Cos²A + Cot²ACos²A.

⇒ Cos²A + Cos²A.Cos²A/Sin²A.

Taking L.C.M in equation, we get.

⇒ Cos²A.Sin²A + Cos⁴ A/Sin²A.

⇒ Cos²A(Sin²A + Cos²A)/Sin²A.

As we know that,

⇒ Sin²∅ + Cos²∅ = 1.

⇒ Cos²A/Sin²A.

⇒ Cot²A.

Hence, L.H.S = R.H.S.

(4) = Sin²∅ + Sin²∅.Cot²∅ = 1.

As we know that,

⇒ Cot²∅ = Cosec²∅ - 1.

⇒ Sin²∅ + Sin²∅(Cosec²∅ - 1).

⇒ Sin²∅ + Sin²∅.Cosec²∅ - Sin²∅.

⇒ Cosec²∅ = 1/Sin²∅.

⇒ Sin²∅ + 1 - Sin²∅.

⇒ 1.

Hence, L.H.S = R.H.S.

(5) = Sin∅(1 + Cot²∅) = Cosec∅.

As we know that,

⇒ 1 + Cot²∅ = Cosec²∅.

⇒ Sin∅(Cosec²∅).

⇒ Cosec²∅ = 1/Sin²∅.

⇒ Sin∅/Sin²∅.

⇒ 1/Sin∅.

⇒ Cosec∅.

Hence, L.H.S = R.H.S.

(6) = Cos A(1 + Tan²A) = Sec A.

As we know that,

⇒ 1 + Tan²∅ = Sec²∅.

⇒ Cos A(Sec²A).

⇒ Sec²A = 1/Cos²A.

⇒ Cos A/Cos²A.

⇒ 1/Cos A.

⇒ Sec A.

Hence, L.H.S = R.H.S.

(7) = (Sin x - Cos x)² = 1 - 2Sinx.Cosx.

As we know that,

⇒ (a - b)² = a² + b² - 2ab.

⇒ (Sin²x + Cos²x - 2Sinx.Cosx).

⇒ Sin²x + Cos²x = 1.

⇒ 1 - 2Sinx.Cosx.

Hence, L.H.S = R.H.S.

(8) = (1 - Sin²A)Cosec²A = Cot²A.

As we know that,

⇒ 1 - Sin²∅ = Cos²∅.

⇒ Cos²A.Cosec²A.

⇒ Cosec²A = 1/Sin²A.

⇒ Cos²A/Sin²A.

⇒ Cot²A.

Hence, L.H.S = R.H.S.

(9) = Cos∅.√1 + Cot²∅ = √Cosec²∅ - 1.

As we know that,

⇒ 1 + Cot²∅ = Cosec²∅.

⇒ Cosec²∅ - 1 = Cot²∅.

⇒ Cos∅.√Cosec²∅. = √Cot²∅.

⇒ Cos∅ Cosec∅ = Cot∅.

⇒ Cosec∅ = 1/Sin∅.

⇒ Cot∅ = Cos∅/Sin∅.

⇒ Cos∅/Sin∅ = Cos∅/Sin∅.

⇒ Cot∅ = Cot∅.

Hence, L.H.S = R.H.S.

(10) = Cos∅ Cosec∅.√Sec²∅ - 1 = 1.

As we know that,

⇒ Sec²∅ - 1 = Tan²∅.

⇒ Cos∅ Cosec∅ .√Tan²∅.

⇒ Cos∅ Cosec∅. Tan∅.

⇒ Cot∅ Tan∅.

⇒ Cot∅ = Cos∅/Sin∅.

⇒ Tan∅ = Sin∅/Cos∅.

⇒ Cos∅/Sin∅ X Sin∅/Cos∅.

⇒ 1.

Hence, L.H.S = R.H.S.

Answer 2

$\bf \:\large \red{AηsωeR : a} ✍$

$\bf \:LHS \: {cos}^{2} θ - {cos}^{2} θ \times {sin}^{2} θ$

$\bf \:  ⟼ {cos}^{2} θ(1 - {sin}^{2} θ)$

$\bf \:  ⟼ {cos}^{2} θ \times {cos}^{2} θ$

$\bf \:  ⟼ {cos}^{4} θ$

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

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$\bf \:\large \red{AηsωeR : b} ✍$

$\bf \:LHS\ ⟼(1 - {cos}^{2}θ)(1 + {tan}^{2} θ)$

$\bf \:  ⟼ {sin}^{2} θ \times {sec}^{2} θ$

$\bf \:  ⟼ \dfrac{ {sin}^{2} θ}{ {cos}^{2}θ }$

$\bf \:  ⟼ {tan}^{2} θ$

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

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$\bf \:\large \red{AηsωeR : c} ✍$

$\bf \:LHS ⟼(1 + {cot}^{2} A)(1 - {sin}^{2} A)$

$\bf \:  ⟼ {cosec}^{2} A \times {cos}^{2} A$

$\bf \:  ⟼ \dfrac{ {cos}^{2}A }{ {sin}^{2}A }$

$\bf \:  ⟼ {cot}^{2} A$

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

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$\bf \:\large \red{AηsωeR : d} ✍$

$\bf \:LHS ⟼ {sin}^{2} θ + {sin}^{2} θ \times {cot}^{2} θ$

$\bf \:  ⟼ {sin}^{2} θ(1 + {cot}^{2} θ)$

$\bf \:  ⟼ {sin}^{2} θ \times {cosec}^{2}θ$

$\bf \:  ⟼ {sin}^{2} θ \times \dfrac{1}{ {sin}^{2}θ } = 1$

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

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$\bf \:\large \red{AηsωeR : e} ✍$

$\bf \:LHS ⟼sinθ(1 + {cot}^{2} θ)$

$\bf \:  ⟼ sinθ \times {cosec}^{2} θ$

$\bf \:  ⟼ sinθ \times \dfrac{1}{ {sin}^{2} θ}$

$\bf \:  ⟼ \dfrac{1}{sinθ} = cosecθ$

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

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$\bf \:\large \red{AηsωeR : f} ✍$

$\bf \:LHS ⟼cosA(1 + {tan}^{2} A)$

$\bf \:  ⟼ cosA \times {sec}^{2} A$

$\bf \:  ⟼ cosA \times \dfrac{1}{ {cos}^{2} A}$

$\bf \:  ⟼ \dfrac{1}{cosA} = secA$

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

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$\bf \:\large \red{AηsωeR : g} ✍$

$\bf \:LHS ⟼ {(sinx - cosx)}^{2}$

$\bf \:  ⟼ {sin}^{2} x + {cos}^{2} x - 2 \times sinx \times cosx$

$\bf \:  ⟼ 1 - 2sinx \: cosx$

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

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$\bf \:\large \red{AηsωeR : h} ✍$

$\bf \:LHS ⟼(1 - {sin}^{2} A) {cosec}^{2} A$

$\bf \:  ⟼ {cos}^{2} A \times \dfrac{1}{ {sin}^{2} A} = {cot}^{2} A$

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

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$\bf \:\large \red{AηsωeR : i} ✍$

$\bf \:LHS ⟼cosθ \times \sqrt{1 + {cot}^{2} θ}$

$\bf \:  ⟼ cosθ \times \sqrt{ {cosec}^{2}θ }$

$\bf \:  ⟼ cosθ \times cosecθ$

$\bf \:  ⟼ cosθ \times \dfrac{1}{sinθ}$

$\bf \:  ⟼ cotθ = \sqrt{ {cot}^{2} θ}$

$\bf \:  ⟼ \sqrt{ {cosec}^{2} θ - 1}$

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

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$\bf \:\large \red{AηsωeR : j} ✍$

$\bf \:LHS ⟼cosθ \: cosecθ \: \sqrt{ {sec}^{2}θ - 1 }$

$\bf \:  ⟼ cosθ \times \dfrac{1}{sinθ} \times \sqrt{ {tan}^{2} θ}$

$\bf \:  ⟼ cotθ \times tanθ = 1$

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

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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

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