9 Most SAQ’s of Inverse Trigonometric Functions Chapter in Inter 1st Year Maths-1A (TS/AP)

4 Marks

SAQ-1 : P.T. sin^-14/5 + sin^-17/25 = sin^-1117/125.

Given:

$$\sin^{-1} \left(\frac{4}{5}\right) = \alpha$$ implies $$\sin \alpha = \frac{4}{5}$$

$$\sin^{-1} \left(\frac{7}{25}\right) = \beta$$ implies $$\sin \beta = \frac{7}{25}$$

To find the cosines of α and β, we use the identity $$\sin^2 \theta + \cos^2 \theta = 1$$

For α : $$\sin^2 \alpha + \cos^2 \alpha = 1 \Rightarrow \left(\frac{4}{5}\right)^2 + \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = 1 – \left(\frac{16}{25}\right) = \frac{9}{25} \Rightarrow \cos \alpha = \frac{3}{5}$$

For β : $$\sin^2 \beta + \cos^2 \beta = 1 \Rightarrow \left(\frac{7}{25}\right)^2 + \cos^2 \beta = 1 \Rightarrow \cos^2 \beta = 1 – \left(\frac{49}{625}\right) = \frac{576}{625} \Rightarrow \cos \beta = \frac{24}{25}$$

Now, using the identity for $$\sin(\alpha + \beta)$$

$$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$

Substituting the given values:

$$\sin(\alpha + \beta) = \left(\frac{4}{5}\right) \left(\frac{24}{25}\right) + \left(\frac{3}{5}\right) \left(\frac{7}{25}\right)$$

$$\sin(\alpha + \beta) = \frac{96}{125} + \frac{21}{125} = \frac{117}{125}$$

Hence, we have shown that:

$$\sin(\alpha + \beta) = \frac{117}{125}$$

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SAQ-2 : P.T. sin^-13/5 + cos^-112/13 = cos^-133/65.

Given:

$$\sin^{-1} \left(\frac{3}{5}\right) = α$$ implies $$\sin α = \frac{3}{5}$$

$$\cos^{-1} \left(\frac{12}{13}\right) = β$$ implies $$\cos β = \frac{12}{13}$$

To find the cosines of α and the sines of β, we use the Pythagorean identities for sine and cosine:

For α :

$$\sin^2 α + \cos^2 α = 1 \Rightarrow \left(\frac{3}{5}\right)^2 + \cos^2 α = 1$$

For β :

$$\sin^2 β + \cos^2 β = 1 \Rightarrow \sin^2 β + \left(\frac{12}{13}\right)^2 = 1$$

Solving for cosα and sinβ :

$$\cos α = \frac{4}{5}$$ since $$1 – \left(\frac{3}{5}\right)^2 = \left(\frac{4}{5}\right)^2$$

$$\sin β = \frac{5}{13}$$ since $$1 – \left(\frac{12}{13}\right)^2 = \left(\frac{5}{13}\right)^2$$

Now, using the identity for $$\cos(α + β)$$

$$\cos(α + β) = \cos α \cos β – \sin α \sin β$$

Substituting the given values:

$$\cos(α + β) = \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) – \left(\frac{3}{5}\right) \left(\frac{5}{13}\right)$$

$$\cos(α + β) = \frac{48}{65} – \frac{15}{65} = \frac{33}{65}$$

Thus, we have shown that:

$$\cos(α + β) = \frac{33}{65}$$

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SAQ-3 : Prove that cos^-14/5 + sin^-13/√34 = Tan^-127/11.

Given:

$$\cos^{-1} \left(\frac{4}{5}\right) = \alpha$$ $$\cos \alpha = \frac{4}{5}$$

$$\sin^{-1} \left(\frac{3}{\sqrt{34}}\right) = \beta$$ $$\sin \beta = \frac{3}{\sqrt{34}}$$

$$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan \alpha \tan \beta}$$

Substitute the given values into the formula:

$$\tan(\alpha + \beta) = \frac{\frac{3}{4} + \frac{3}{5}}{1 – \left(\frac{3}{4} \cdot \frac{3}{5}\right)}$$

Let’s calculate the exact value:

$$\tan(\alpha + \beta) = \frac{\frac{15}{20} + \frac{12}{20}}{1 – \frac{9}{20}}$$

$$\tan(\alpha + \beta) = \frac{27}{20 – 9}$$

$$\tan(\alpha + \beta) = \frac{27}{11}$$

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SAQ-4 : P.T. Tan^-11/2 + Tan^-11/5 + Tan^-11/8 = π/4.

We know, $$\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{x + y}{1 – xy} \right)$$

$$\tan^{-1} \left(\frac{1}{2}\right) + \tan^{-1} \left(\frac{1}{5}\right)$$

$$\tan^{-1} \left( \frac{\frac{1}{2} + \frac{1}{5}}{1 – \frac{1}{2} \cdot \frac{1}{5}} \right)$$

Let’s calculate the exact value of this expression:

$$\tan^{-1} \left( \frac{\frac{5}{10} + \frac{2}{10}}{1 – \frac{1}{10}} \right) = \tan^{-1} \left( \frac{7}{10 – 1} \right) = \tan^{-1} \left( \frac{7}{9} \right)$$

So, the correct equation up to this point is:

$$\tan^{-1} \left(\frac{1}{2}\right) + \tan^{-1} \left(\frac{1}{5}\right) = \tan^{-1} \left( \frac{7}{9} \right)$$

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SAQ-5 : P.T. Tan^-13/4 + Tan^-13/5 – Tan^-18/19 = π/4.

$$\tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{3}{5}\right)$$

$$\tan^{-1}\left(\frac{3}{4} + \frac{3}{5}\right) / \left(1 – \frac{3}{4} \cdot \frac{3}{5}\right)$$

$$\tan^{-1}\left(\frac{15}{20} + \frac{12}{20}\right) / \left(\frac{20}{20} – \frac{9}{20}\right)$$

$$\tan^{-1}\left(\frac{27}{11}\right)$$

$$\tan^{-1}\left(\frac{27}{11}\right) – \tan^{-1}\left(\frac{8}{19}\right)$$

$$\tan^{-1}\left(\frac{27}{11} – \frac{8}{19}\right) / \left(1 + \frac{27}{11} \cdot \frac{8}{19}\right)$$

$$\tan^{-1}\left(\frac{513 – 88}{209 + 216}\right)$$

$$\tan^{-1}\left(\frac{425}{425}\right)$$

$$\tan^{-1}(1) = \frac{\pi}{4}$$

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SAQ-6 : Prove that sin^-14/5 + 2Tan^-11/3 = π/2.

$$2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$$

$$2\tan^{-1}\left(\frac{1}{3}\right) = \cos^{-1}\left(\frac{1-\left(\frac{1}{3}\right)^2}{1+\left(\frac{1}{3}\right)^2}\right) = \cos^{-1}\left(\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\right)$$

$$= \cos^{-1}\left(\frac{8/9}{10/9}\right) = \cos^{-1}\left(\frac{8}{10}\right) = \cos^{-1}\left(\frac{4}{5}\right)$$

$$\sin^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{4}{5}\right) = \pi/2$$

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SAQ-7 : P.T. cos(2Tan^-11/7) = sin(2Tan^-13/4).

$$\tan^{-1}\left(\frac{1}{7}\right) = \alpha$$ implies $$\tan(\alpha) = \frac{1}{7}$$

$$\tan^{-1}\left(\frac{3}{4}\right) = \beta$$ implies $$\tan(\beta) = \frac{3}{4}$$

$$\cos(2\tan^{-1}\left(\frac{1}{7}\right)) = \sin(2\tan^{-1}\left(\frac{3}{4}\right))$$

Using the identities:

$$\cos(2\alpha) = \frac{1-\tan^2(\alpha)}{1+\tan^2(\alpha)}$$

$$\sin(2\beta) = \frac{2\tan(\beta)}{1+\tan^2(\beta)}$$

After performing the calculations:

The left-hand side $$\cos(2\tan^{-1}\left(\frac{1}{7}\right)) = 0.96$$

The right-hand side $$\sin(2\tan^{-1}\left(\frac{3}{4}\right)) = 0.96$$

$$\cos(2\tan^{-1}\left(\frac{1}{7}\right)) = \sin(2\tan^{-1}\left(\frac{3}{4}\right))$$

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SAQ-8 : P.T. cos(2Tan^-11/7) = sin(4Tan^-11/3).

$$\tan^{-1}\left(\frac{1}{7}\right) = \alpha$$ implies $$\tan(\alpha) = \frac{1}{7}$$

$$\tan^{-1}\left(\frac{1}{3}\right) = \beta$$ implies $$\tan(\beta) = \frac{1}{3}$$

LHS Calculation:

$$\cos(2\tan^{-1}\left(\frac{1}{7}\right)) = \cos(2\alpha)$$

$$\cos(2\alpha) = \frac{1 – \tan^2(\alpha)}{1 + \tan^2(\alpha)}$$

$$\tan(\alpha) = \frac{1}{7}$$

$$\frac{1 – \left(\frac{1}{7}\right)^2}{1 + \left(\frac{1}{7}\right)^2} = \frac{1 – \frac{1}{49}}{1 + \frac{1}{49}} = \frac{48}{50} = \frac{24}{25}$$

RHS Calculation:

$$\sin(4\tan^{-1}\left(\frac{1}{3}\right)) = \sin(4\beta)$$

$$\tan(2\beta) = \frac{2\tan(\beta)}{1 – \tan^2(\beta)}$$

$$\tan(\beta) = \frac{1}{3}$$

$$\tan(2\beta) = \frac{2\cdot\frac{1}{3}}{1 – \left(\frac{1}{3}\right)^2} = \frac{2/3}{8/9} = \frac{3}{4}$$

$$\sin(4\beta) = \frac{2\tan(2\beta)}{1 + \tan^2(2\beta)}$$

$$\frac{2\cdot\frac{3}{4}}{1 + \left(\frac{3}{4}\right)^2} = \frac{3/2}{1 + 9/16} = \frac{24}{25}$$

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SAQ-9 : If Tan^-1x + Tan^-1y + Tan^-1z = π/2, then prove that xy+yz+zx=1.

Given:

Let $$\tan^{-1}(x) = A$$ implies $$\tan(A) = x$$

$$\tan^{-1}(y) = B$$ implies $$\tan(B) = y$$

$$\tan^{-1}(z) = C$$ implies $$\tan(C) = z$$

Hence, $$A + B + C = \frac{\pi}{2}$$

$$A + B = \frac{\pi}{2} – C$$

$$\tan(A + B) = \tan\left(\frac{\pi}{2} – C\right)$$

$$\frac{\tan(A) + \tan(B)}{1 – \tan(A)\tan(B)} = \frac{1}{\tan(C)}$$

$$\frac{x + y}{1 – xy} = \frac{1}{z}$$

$$zx + yz = 1 – xy$$

$$xy + yz + zx = 1$$