Question

Sum number 24 and 26.

Answer 1

Answer:

For sum 24

Step-by-step explanation:

Write expression in terms of sin and cos

LHS=(sinA/cosA)/(1-cosA/sinA)+(cosA/sinA)/(1-sinA/cosA)

=(sinA/cosA)(sinA-cosA/sinA)+(cosA/sinA)/(cosA-sinA/cosA)

=(sin^2A/cosA(sinA-cosA))+(cos^2A/sinA(cosA-sinA))

=(sin^2A/cosA(sinA-cosA))-(cos^2A/sinA(sinA-cosA))

(Taken -ve to make it easy to take LCM)

Now taking LCM,we hv to mulptiply sinA to left expression and cosA to the right expression to get the common denominator cosAsinA(sinA-cosA)

We get LHS=(sin^2A(sinA)-cos^2A(cosA))/cosAsinA(sinA-cosA)

=sin^3A-cos^3A/(sinA-cosA)cosAsinA

If u are in 10th std make sure to remember ur formulas of a^3-b^3,a^3+b^3,(a+b)^3 and (a-b)^3

a^3-b^3=(a-b)(a^2+b^2+ab)

We hv to apply this formula to the numerator.

We get (sinA-cosA)(sin^2A+cos^2A+sinAcosA)/cosAsinA(sinA-cosA)

(sinA-cosA) gets cancelled

LHS=(sin^2A+cos^2A+sinAcosA)/cosAsinA

=(1+sinAcosA)/cosAsinA

=(1/sinAcosA)+(sinAcosA/cosAsinA)

=cosecAsecA+1

I am sorry if the answer looks confusing.I am not familiar with how answers are written in code.

Answer 2

Question 24 - To prove:

$\longrightarrow \ \sf \dfrac{tanA}{1 - cotA} + \dfrac{cotA}{1 - tanA} = secA + cosecA + 1$

Taking the LHS we get;

$\longrightarrow \ \sf \dfrac{tanA}{1 - cotA} + \dfrac{cotA}{1 - tanA}$

The RHS is in terms of secA and cosecA, so let's try to express the LHS in terms of secA and cosecA as well.

We can do that by using;

➝ tanA = sinA/cosA

➝ cotA = cosA/sinA

$\longrightarrow \ \sf \dfrac{\bigg[\dfrac{sinA}{cosA}\bigg]}{1 - \bigg[\dfrac{cosA}{sinA}\bigg]} + \dfrac{\bigg[\dfrac{cosA}{sinA}\bigg]}{1 - \bigg[\dfrac{sinA}{cosA}\bigg]}$

Taking LCM we get;

$\longrightarrow \ \sf \dfrac{\bigg[\dfrac{sinA}{cosA}\bigg]}{\bigg[\dfrac{sinA - cosA}{sinA}\bigg]} + \dfrac{\bigg[\dfrac{cosA}{sinA}\bigg]}{\bigg[\dfrac{cosA - sinA}{cosA}\bigg]}$

$\longrightarrow \ \sf \bigg[\dfrac{sinA}{cosA} \times \dfrac{sinA}{sinA - cosA}\bigg] + \bigg[\dfrac{cosA}{sinA} \times \dfrac{cosA}{cosA - sinA}\bigg]$

$\longrightarrow \ \sf \dfrac{sinA \times sinA}{cosA \big[sinA - cosA\big]} + \dfrac{cosA \times cosA}{sinA \big[cosA - sinA\big]}$

$\longrightarrow \ \sf \dfrac{sin^2A}{cosA \big[sinA - cosA\big]} + \dfrac{cos^2A}{sinA \big[cosA - sinA\big]}$

Using 1/(a - b) = -[1/(b - a)] we get;

[Same as a - b = -(b - a)]

$\longrightarrow \ \sf \dfrac{sin^2A}{cosA \big[sinA - cosA\big]} - \dfrac{cos^2A}{sinA \big[sinA - cosA\big]}$

Take 1/sinA - cosA as common.

$\longrightarrow \ \sf \dfrac{1}{sinA - cosA} \ \Bigg[\dfrac{sin^2A}{cosA} - \dfrac{cos^2A}{sinA} \Bigg]$

Taking LCM we get;

$\longrightarrow \ \sf \dfrac{1}{sinA - cosA} \ \Bigg[\dfrac{sin^3A - cos^3A}{cosAsinA} \Bigg]$

Using a³ - b³ = (a - b)(a² + ab + b²) we get:

• Where a = sin³A & b = cos³A.

$\longrightarrow \ \sf \dfrac{1}{sinA - cosA} \ \Bigg[\dfrac{\big(sinA - cosA\big) \big(sin^2A + sinAcosA + cos^2A\big)}{cosAsinA} \Bigg]$

On cancelling (sinA - cosA) and using sin²A + cos²A = 1 we get:

$\longrightarrow \ \sf \dfrac{1 + sinAcosA}{cosAsinA}$

$\longrightarrow \ \sf \dfrac{1}{cosAsinA} + \dfrac{sinAcosA}{cosAsinA}$

$\longrightarrow \ \sf \dfrac{1}{cosA} \times \dfrac{1}{sinA} + \dfrac{sinAcosA}{cosAsinA}$

Using the following we get:

• 1/cosA = secA

• 1/sinA = cosecA

• Cancelling sinAcosA and cosAsinA

$\longrightarrow \ \sf secA \times cosecA} + 1$

LHS = RHS

Hence Proved.

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Question 26 - To prove:

$\longrightarrow \sf \dfrac{cot\theta + cosec\theta - 1}{cot\theta - cosec\theta + 1} = \dfrac{1 + cos\theta}{sin\theta}$

Taking the LHS we get;

$\longrightarrow \sf \dfrac{cot\theta + cosec\theta - 1}{cot\theta - cosec\theta + 1}$

Using 1 = cosec²θ - cot²θ we get:

$\longrightarrow \sf \dfrac{cot\theta + cosec\theta - \big(cosec^2\theta - cot^2\theta\big)}{cot\theta - cosec\theta + 1}$

Using a² - b² = (a + b)(a - b) we get:

$\longrightarrow \sf \dfrac{cot\theta + cosec\theta - \big(cosec\theta + cot\theta\big)\big(cosec\theta - cot\theta\big)}{cot\theta - cosec\theta + 1}$

Taking cotθ + cosecθ out as common from the numerator we get;

$\longrightarrow \sf \dfrac{cot\theta + cosec\theta\Big[1 - \big(cosec\theta - cot\theta\big)\Big]}{cot\theta - cosec\theta + 1}$

$\longrightarrow \sf \dfrac{cot\theta + cosec\theta\Big[1 - cosec\theta + cot\theta\Big]}{1 - cosec\theta + cot\theta}$

Canceling [1 - cosecθ + cotθ] in the numerator and denominator we get;

$\longrightarrow \sf cot\theta + cosec\theta$

Using the below ratios we get;

• cotθ = cosθ/sinθ

• cosecθ = 1/sinθ

$\longrightarrow \sf \dfrac{cos\theta}{sin\theta} + \dfrac{1}{sin\theta}$

$\longrightarrow \sf \dfrac{1 +cos\theta}{sin\theta}$

LHS = RHS

Hence proved.