CBSE Class 11 Maths Exemplar Chapter 3: Trigonometric Functions NCERT Solutions

We are asked to prove the trigonometric identity:

\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{\cos A} . Let's start with the Left Hand Side (LHS):

\nLHS = \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}

We know the fundamental trigonometric identity

\sec^2 A - \tan^2 A = 1

. We can use this to substitute for '1' in the numerator. Also, recall that

\sec^2 A - \tan^2 A = (\sec A + \tan A)(\sec A - \tan A) . \nLHS = \frac{\tan A + \sec A - (\sec^2 A - \tan^2 A)}{\tan A - \sec A + 1}

Now, factor the term

\sec^2 A - \tan^2 A

in the numerator:

\nLHS = \frac{(\tan A + \sec A) - (\sec A + \tan A)(\sec A - \tan A)}{(\tan A - \sec A + 1)}

Factor out

(\sec A + \tan A)

from the numerator:

\nLHS = \frac{(\sec A + \tan A)(1 - (\sec A - \tan A))}{1 - \sec A + \tan A}

Notice that

1 - (\sec A - \tan A) = 1 - \sec A + \tan A

, which is the denominator. So, we can cancel this term:

\nLHS = \frac{(\sec A + \tan A)(\tan A - \sec A + 1)}{1 - \sec A + \tan A}

\nLHS = \sec A + \tan A

Now, express

\sec A and \tan A

in terms of

\sin A and \cos A : \nLHS = \frac{1}{\cos A} + \frac{\sin A}{\cos A}

Combine the terms over a common denominator:

\nLHS = \frac{1 + \sin A}{\cos A}

This is equal to the Right Hand Side (RHS). Hence proved.

We are given that \ny = \frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha} . We need to prove that

\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}

is also equal to y.

Let's consider the expression

\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}

. To simplify this, we can multiply the numerator and denominator by

(1+\cos\alpha+\sin\alpha)

to relate it to the given expression for y. However, a more direct approach is to manipulate the expression to match the form of y.

Let's try multiplying the numerator and denominator by

(1+\sin\alpha)

to see if it simplifies. This doesn't seem to lead directly to the form of y. Instead, let's try to simplify the expression by using trigonometric identities.

Consider the expression

\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}

. Let's try to manipulate it to match the given y.

Let's re-examine the provided solution's approach. It multiplies the numerator and denominator by

(1+\cos\alpha+\sin\alpha)

. This seems incorrect as it would complicate the denominator.

Let's try a different approach. We can use half-angle formulas. Let

\alpha = 2\theta

. Then

\sin\alpha = 2\sin\theta\cos\theta , \cos\alpha = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1 . The given expression for y becomes:

\ny = \frac{2(2\sin\theta\cos\theta)}{(1 + (2\cos^2\theta - 1)) + 2\sin\theta\cos\theta} = \frac{4\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta} = \frac{4\sin\theta\cos\theta}{2\cos\theta(\cos\theta + \sin\theta)} = \frac{2\sin\theta}{\cos\theta + \sin\theta}

Now consider the second expression

\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha} : \frac{1-(1-2\sin^2\theta)+2\sin\theta\cos\theta}{1+2\sin\theta\cos\theta} = \frac{2\sin^2\theta+2\sin\theta\cos\theta}{1+2\sin\theta\cos\theta}

This doesn't seem to simplify easily to match y.

Let's follow the original solution's steps carefully, assuming there might be a simplification we missed.

Given \ny = \frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}

Consider the expression

\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}

. The provided solution multiplies by

\frac{1+\cos\alpha+\sin\alpha}{1+\cos\alpha+\sin\alpha}

. Let's re-evaluate this step.

\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha} \times \frac{1+\cos\alpha+\sin\alpha}{1+\cos\alpha+\sin\alpha}

Numerator:

((1+\sin\alpha)-\cos\alpha)((1+\sin\alpha)+\cos\alpha) = (1+\sin\alpha)^2 - \cos^2\alpha

= (1 + 2\sin\alpha + \sin^2\alpha) - \cos^2\alpha

Using

\cos^2\alpha = 1 - \sin^2\alpha : = 1 + 2\sin\alpha + \sin^2\alpha - (1 - \sin^2\alpha)

= 1 + 2\sin\alpha + \sin^2\alpha - 1 + \sin^2\alpha

= 2\sin^2\alpha + 2\sin\alpha = 2\sin\alpha(\sin\alpha + 1)

Denominator:

(1+\sin\alpha)(1+\cos\alpha+\sin\alpha)

So the expression becomes:

\frac{2\sin\alpha(\sin\alpha + 1)}{(1+\sin\alpha)(1+\cos\alpha+\sin\alpha)}

Cancel out

(1+\sin\alpha) : = \frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}

This is exactly the expression for y.

Hence proved.

We are given the equation

m\sin\theta = n\sin(\theta + 2\alpha) . Rearrange the equation to form a ratio:

\frac{\sin(\theta + 2\alpha)}{\sin \theta} = \frac{m}{n}

Now, we apply the componendo and dividendo rule. The rule states that if

\frac{a}{b} = \frac{c}{d}

, then

\frac{a+b}{a-b} = \frac{c+d}{c-d} . Applying this to our ratio:

\frac{\sin(\theta + 2\alpha) + \sin \theta}{\sin(\theta + 2\alpha) - \sin \theta} = \frac{m + n}{m - n}

We use the sum-to-product and difference-to-product trigonometric formulas:

\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)

\sin x - \sin y = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)

Applying these formulas to the numerator and denominator of the LHS:

\frac{2\sin\left(\frac{(\theta+2\alpha)+\theta}{2}\right)\cos\left(\frac{(\theta+2\alpha)-\theta}{2}\right)}{2\cos\left(\frac{(\theta+2\alpha)+\theta}{2}\right)\sin\left(\frac{(\theta+2\alpha)-\theta}{2}\right)} = \frac{m+n}{m-n}

Simplify the terms inside the trigonometric functions:

\frac{2\sin\left(\frac{2\theta+2\alpha}{2}\right)\cos\left(\frac{2\alpha}{2}\right)}{2\cos\left(\frac{2\theta+2\alpha}{2}\right)\sin\left(\frac{2\alpha}{2}\right)} = \frac{m+n}{m-n}

\frac{\sin(\theta + \alpha)\cos \alpha}{\cos(\theta + \alpha)\sin \alpha} = \frac{m+n}{m-n}

Recognize that

\frac{\sin x}{\cos x} = \tan x and \frac{\cos x}{\sin x} = \cot x . \tan(\theta + \alpha) \cdot \cot \alpha = \frac{m+n}{m-n}

Hence proved.

We are given

\cos(\alpha + \beta) = \frac{4}{5} and \sin(\alpha - \beta) = \frac{5}{13} . Since

\alpha and \beta

are acute angles (

0 < \alpha, \beta < \frac{\pi}{2}

), it implies that

\alpha + \beta and \alpha - \beta

are within certain ranges. Specifically,

0 < \alpha + \beta < \pi and -\frac{\pi}{2} < \alpha - \beta < \frac{\pi}{2} . From

\cos(\alpha + \beta) = \frac{4}{5}

, we can find

\sin(\alpha + \beta)

using the identity

\sin^2 x + \cos^2 x = 1 . \sin^2(\alpha + \beta) = 1 - \cos^2(\alpha + \beta) = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}

\sin(\alpha + \beta) = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}

Since

0 < \alpha + \beta < \pi , \sin(\alpha + \beta)

can be positive. Given the context of typical problems, we assume

\sin(\alpha + \beta) = \frac{3}{5} . From

\sin(\alpha - \beta) = \frac{5}{13}

, we can find

\cos(\alpha - \beta) . \cos^2(\alpha - \beta) = 1 - \sin^2(\alpha - \beta) = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}

\cos(\alpha - \beta) = \pm \sqrt{\frac{144}{169}} = \pm \frac{12}{13}

Since

-\frac{\pi}{2} < \alpha - \beta < \frac{\pi}{2} , \cos(\alpha - \beta)

must be positive. Therefore,

\cos(\alpha - \beta) = \frac{12}{13} . We need to find

\tan 2\alpha

. We know that

2\alpha = (\alpha + \beta) + (\alpha - \beta) . First, let's find

\tan(\alpha + \beta) and \tan(\alpha - \beta) . \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{3/5}{4/5} = \frac{3}{4}

\tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{5/13}{12/13} = \frac{5}{12}

Now, we use the tangent addition formula:

\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} . Let

x = \alpha + \beta and y = \alpha - \beta

. Then

x+y = 2\alpha . \tan(2\alpha) = \tan((\alpha + \beta) + (\alpha - \beta)) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta)\tan(\alpha - \beta)}

Substitute the values we found:

\tan(2\alpha) = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}

Find a common denominator for the numerator:

\frac{3}{4} + \frac{5}{12} = \frac{9}{12} + \frac{5}{12} = \frac{14}{12} = \frac{7}{6}

Calculate the denominator:

1 - \left(\frac{3}{4}\right)\left(\frac{5}{12}\right) = 1 - \frac{15}{48} = 1 - \frac{5}{16} = \frac{16 - 5}{16} = \frac{11}{16}

Now, divide the numerator by the denominator:

\tan(2\alpha) = \frac{\frac{7}{6}}{\frac{11}{16}} = \frac{7}{6} \times \frac{16}{11} = \frac{7 \times 8}{3 \times 11} = \frac{56}{33}

Thus, the value of

\tan 2\alpha is \frac{56}{33} .