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Maths 2 Marks Questions

P-2 Trigonometry · Maharashtra Board STD 10 (MSBSHSE - English Medium). 24 questions.

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Question 12 Marks
If $\sin \theta=\frac{11}{61}$ find the values of cosθ using trigonometric identity.
Answer
As,
$\sin ^2 \theta+\cos ^2 \theta=1$
$\Rightarrow\left(\frac{11}{61}\right)^2+\cos ^2 \theta=1$
$\Rightarrow \frac{121}{3721}+\cos ^2 \theta=1$
$\Rightarrow \cos ^2 \theta=1-\frac{121}{3721}=\frac{3600}{3721}$
$\Rightarrow \cos \theta=\frac{60}{61}$
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Question 22 Marks
Prove the following.
$\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=2 \sec ^2 \theta$
Answer
Taking LHS
$\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}$
$=\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{2}{1-\sin ^2 \theta}$
$=\frac{2}{\cos ^2 \theta}\left[\sin ^2 \theta+\cos ^2 \theta=1\right]$
$=2 \sec ^2 \theta$
= RHS
= Proved
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Question 32 Marks
Prove that:
$\frac{\sin ^2 \theta}{\cos \theta}+\cos \theta=\sec \theta$
Answer
Taking LHS
$\frac{\sin ^2 \theta}{\cos \theta}+\cos \theta$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}\left[ As , \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\sec \theta\left[ As , \sec \theta=\frac{1}{\cos \theta}\right]$
= RHS
Proved !
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Question 42 Marks
Prove the following.
$tan^4\theta + tan^2\theta = sec^4\theta - sec^2\theta$
 
Answer
Taking LHS
$tan^4\theta + tan^2\theta$
$= tan^2\theta ( tan^2\theta + 1)$
$= (sec^2\theta - 1)(sec^2\theta ) [1 + tan^2\theta = sec^2\theta ]$
$= sec^4\theta - sec^2\theta$
= RHS
Proved !
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Question 52 Marks
$\sin \theta=\frac{7}{25}$ find the values of cosθ and tanθ.
Answer
We know that,
$\sin ^2 \theta+\cos ^2 \theta=1$
$\Rightarrow\left(\frac{7}{25}\right)^2+\cos ^2 \theta=1$
$\Rightarrow \frac{49}{625}+\cos ^2 \theta=1$
$\Rightarrow \cos ^2 \theta=1-\frac{49}{625}$
$\Rightarrow \cos ^2 \theta=\frac{576}{625}$
$\Rightarrow \cos \theta=\frac{24}{25}$
Also,
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \tan \theta=\frac{\left(\frac{7}{25}\right)}{\frac{24}{25}}=\frac{7}{24}$
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Question 62 Marks
Prove the following.
$cot^2\theta - tan^2\theta = cosec^2\theta - sec^2\theta $
 
Answer
Taking LHS
$cot^2\theta - tan^2\theta $
$[ Now, cosec^2\theta - 1 = cot^2\theta and sec^2\theta - 1 = tan^2\theta ]$
$= cosec^2\theta - 1 - (sec^2\theta - 1)$
$= cosec^2\theta - sec^2\theta $
= RHS
Proved !
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Question 72 Marks
Show that $\sec x+\tan x=\sqrt{\frac{1+\sin x}{1-\sin x}}$
Answer
$\begin{aligned}
\sec x+\tan x & =\frac{1}{\cos x}+\frac{\sin x}{\cos x} \\
& =\frac{1+\sin x}{\cos x} \\
& =\sqrt{\frac{(1+\sin x)^2}{\cos ^2 x}} \\
& =\sqrt{\frac{(1+\sin x)(1+\sin x)}{1-\sin ^2 x}} \\
& =\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}} \\
& =\sqrt{\frac{1+\sin x}{1-\sin x}}
\end{aligned}
$
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Question 82 Marks
$\cos \theta=\frac{\sqrt{3}}{2}$ then find the value of $\frac{1-\sec \theta}{1+\operatorname{cosec} \theta}$.
Answer
$ \cos \theta=\frac{\sqrt{3}}{2} \quad \therefore \sec \theta=\frac{2}{\sqrt{3}}$
$\sin ^2 \theta+\cos ^2 \theta=1$
$\therefore \sin ^2 \theta+\left(\frac{\sqrt{3}}{2}\right)^2=1$
$\therefore \sin ^2 \theta=1-\frac{3}{4}=\frac{1}{4}$
$\therefore \sin \theta=\frac{1}{2} \quad \therefore \operatorname{cosec} \theta=2$
$\therefore \frac{1-\sec \theta}{1+\operatorname{cosec} \theta}=\frac{1-\frac{2}{\sqrt{3}}}{1+2}$
$=\frac{\frac{\sqrt{3}-2}{\sqrt{3}}}{3}$
$=\frac{\sqrt{3}-2}{3 \sqrt{3}}$
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Question 92 Marks
If $\sec \theta=\frac{25}{7}$, find the value of $\tan \theta$.
Answer
Method I
we have,
$
\begin{aligned}
1+\tan ^2 \theta & =\sec ^2 \theta \\
\therefore \quad 1+\tan ^2 \theta & =\left(\frac{25}{7}\right)^2 \\
\therefore \quad \tan ^2 \theta & =\frac{625}{49}-1 \\
& =\frac{625-49}{49} \\
& =\frac{576}{49} \\
\therefore \quad \tan \theta & =\frac{24}{7}
\end{aligned}
$
Image
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Question 102 Marks
If $\sin \theta=\frac{20}{29}$ then find $\cos \theta$
Answer
Method I
We have
$
\sin ^2 \theta+\cos ^2 \theta=1
$
$
\begin{array}{c}
\left(\frac{20}{29}\right)^2+\cos ^2 \theta=1 \\
\frac{400}{841}+\cos ^2 \theta=1 \\
\cos ^2 \theta=1-\frac{400}{841} \\
=\frac{441}{841}
\end{array}
$
Taking square root of both sides.
$
\cos \theta=\frac{21}{29}
$
Image
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Question 112 Marks
A building is $200 \sqrt{3}$ metres high. Find the angle of elevation if its top is 200 m away from its foot.
Answer
60
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Question 202 Marks
$\cos \theta=\frac{\sqrt{3}}{2}$ then find the value of $\frac{1-\sec \theta}{1+\operatorname{cosec} \theta}$.
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Question 242 Marks
A building is $200 \sqrt{3}$ metres high. Find the angle of elevation if its top is 200 m away from its foot.
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