Question

prove the following 1). sin cube teta + cos cube teta divided sin teta +cos teta = 1-sin teta cos teta

Answer 1

Given to prove :-

$\dfrac{sin^3 \theta + cos^3 \theta}{sin\theta + cos\theta} = 1- sin\theta cos\theta$

SOLUTION:-

As we know the formula of a³+ b³

a³+ b³ = (a+b)(a²-ab + b²) By using this we can solve So,

Numerator is sin³θ + cos³θ By using above formula we get

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sin³θ + cos³θ = (sinθ+cosθ) (sin²θ - sinθcosθ + sin²θ)

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So,

$\dfrac{sin^3 \theta + cos^3 \theta}{sin\theta + cos\theta}$

$\dfrac{(sin\theta+cos\theta) (sin^2\theta-sin\theta cos\theta + cos^2\theta)}{(sin\theta + cos\theta)}$

${=} sin^2\theta-sin\theta cos\theta + cos^2\theta$

From trigonometric identities

We know that ,

sin²θ + cos²θ = 1

${=} (sin^2\theta + cos^2\theta- sin\theta cos\theta)$

${=} 1 - sin\theta cos\theta$

So,

$\dfrac{sin^3 \theta + cos^3 \theta}{(sin\theta + cos\theta)} = 1- sin\theta cos\theta$

LHS = RHS

Hence proved !

Used formulae:-

• a³+ b³ = (a+b)(a²-ab + b²)
• sin²θ + cos²θ = 1

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Know more :-

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

Answer 2

Step-by-step explanation:

To Prove :+

sin³θ+cos³θ/sinθ+cosθ = 1 - sinθcosθ

Solution :-

Taking L.H.S :-

= sin³θ+cos³θ/sinθ+cosθ

= (sinθ+cosθ)³ - 3sinθcosθ(sinθ+cosθ)/sinθ+cosθ

[ ∴ a³ + b³ = (a + b)³ - 3ab(a + b)]

Taking "sinθ+cosθ" common :-

= sinθ+cosθ[(sinθ+cosθ)² - 3sinθcosθ]/sinθ+cosθ

= (sinθ+cosθ)² - 3sinθcosθ

= sin²θ + cos²θ + 2sinθcosθ - 3sinθcosθ

= 1 - sinθcosθ

[ ∴ sin²θ + cos²θ = 1]

Know More :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Trigonometric Identities :-

1) sin²A + cos²A = 1

2) sec²A - tan²A = 1

3) csc²A - cot²A = 1