Question

If sin θ = $\sf\dfrac{c}{\sqrt{c^2+d^2}}$ and d > 0, find the value of cos θ & tan θ​

Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• sin θ = $\sf\dfrac{c}{\sqrt{c^2+d^2}}$ and d > 0.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Find the value of cos θ & tan θ.

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• Sin ∅ = $\sf\dfrac{c}{\sqrt{c^2+d^2}}$

We know that,

$\boxed{\pink{\sf sin^{2} \: ∅ \: + \: cos^{2} \: ∅ \: = \: 1}}$

Therefore,

• cos² ∅ = 1 - sin² ∅

Substituting the values,

cos² ∅ = 1 - $\sf(\dfrac{c}{\sqrt{c^2+d^2}})^{2}$

cos² ∅ = 1 - $\sf\dfrac{c^{2}}{c^2+d^2}$

Taking LCM,

cos² ∅ = $\sf\dfrac{c^2+d^2}{c^2+d^2} - \dfrac{c^{2}}{c^2+d^2}$

cos² ∅ = $\sf\dfrac{c^2+d^2 \: - \: c^2}{c^2+d^2}$

cos² ∅ = $\sf\dfrac{\cancel{c^2}+d^2 \: \cancel{- \: c^2}}{c^2+d^2}$

cos² ∅ = $\sf\dfrac{d^2}{c^2+d^2}$

Transposing square to RHS,

cos ∅ = $\sf\sqrt{\dfrac{d^2}{c^2+d^2}}$

cos ∅ = $\sf\dfrac{\sqrt{d^2}}{\sqrt{c^2+d^2}}$

cos ∅ = $\sf\dfrac{d}{\sqrt{c^2+d^2}}$

Therefore,

• cos ∅ = $\sf\dfrac{d}{\sqrt{c^2+d^2}}$

We know that,

• Tan ∅ = $\sf \dfrac{sin \: ∅}{cos \: ∅}$

Substituting the values,

Tan ∅ = $\sf \dfrac{\dfrac{c}{\sqrt{c^2+d^2}}}{\dfrac{d}{\sqrt{c^2+d^2}}}$

Tan ∅ = $\sf \dfrac{\dfrac{c}{\cancel{\sqrt{c^2+d^2}}}}{\dfrac{d}{\cancel{\sqrt{c^2+d^2}}}}$

Tan ∅ = $\sf \dfrac{c}{d}$

Therefore,

• $\underline{\boxed{\therefore{\blue{\sf cos \: ∅\: = \: \dfrac{d}{\sqrt{c^2+d^2}}}}}}$

• $\underline{\boxed{\therefore{\blue{\sf Tan \: ∅ \: = \: \dfrac{c}{d}}}}}$

Answer 2

Step-by-step explanation:

Given:-

Sin θ = c / √(c^2+d^2) ,d > 0

To find:-

Find the values of Cos θ and Tan θ ?

Solution:-

Method -1:-

Given that

Sin θ = c / √(c^2+d^2) ,d > 0---------(1)

On squaring both sides then

=> (Sin θ)^2 =[ c / √(c^2+d^2)]^2

=> Sin^2 θ = c^2/(c^2+d^2)

=> 1- Sin^2 θ = 1-[c^2/(c^2+d^2)]

=> 1- Sin^2 θ =[ (c^2+d^2)-c^2]/(c^2+d^2)

=> 1- Sin^2 θ = d^2/(c^2+d^2)

We know that

Sin^2 A + Cos^2 A = 1

=> Cos^2 θ = d^2/(c^2+d^2)

=> Cos θ =√[d^2/(c^2+d^2)]

Since d > 0

Cos θ = d / √(c^2+d^2) --------(1)

On dividing (1) by (2) then

From (1)&(2) then

=> Sin θ/ Cos θ =[c /√(c^2+d^2)] / [d /√(c^2+d^2)]

=> Sin θ/Cos θ = c/d

Since √(c^2+d^2) is cancelled in both numerator and denominator

We know that

Tan A = Sin A / Cos A

Tan θ = c/d

Method -2:-

Given that

Sin θ = c / √(c^2+d^2) ,d > 0---------(1)

We know that

Sin A = Opposite side to angle A/ Hypotenuse

Opposite side to angle A/ Hypotenuse

= c / √(c^2+d^2)

Let the opposite side = ck units

Let the hypotenuse = √(c^2+d^2) k units

In the right triangle ABC ,

By Pythagoras theorem

AC^2 = AB^2+BC^2

[√(c^2+d^2)k]^2 = AB^2 + (ck)^2

=> (c^2+d^2)k^2-(c^2k^2) = AB^2

=> c^2k^2+d^2k^2-c^2k^2 = AB^2

AB^2 = d^2k^2

AB=√(d^2k^2)

AB = dk units

Now Cos A

=Adjacent side to A/Hypotenuse

=> Cos θ = AB/AC

=> Cos θ = dk/√(c^2+d^2)k

=> Cos θ = d/√(c^2+d^2)

Now Tan A

= Opposite side to A/ Adjacent side to A

Tan θ = BC/AB

=> ck /dk

=> c/d

Tan θ = c/d

Answer:-

The value of Cos θ = d / √(c^2+d^2)

The value of Tan θ = c/d

Used formulae:-

• Sin A = Opposite side to angle A/ Hypotenuse
• Cos A =Adjacent side to A/Hypotenuse
• Tan A = Opposite side to A/ Adjacent side to A
• Tan A = Sin A/ Cos A
• Pythagoras theorem:-

In a right angled triangle , The square of the hypotenuse is equal to the sum of the squares of the other two sides.