Tag: All NCERT Questions solved Maths Class 10th

Q1. Evaluate the following

(i) $\sin {60}^{\circ }\cos {30}^{\circ }+\sin {30}^{\circ }\cos {60}^{\circ }$

Solution
Use the sine addition formula: $\sin (x+y)=\sin x\cos y+\cos x\sin y$
So the expression = $\sin ({60}^{\circ }+{30}^{\circ })=\sin {90}^{\circ }=1.$

Answer: $1$.

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(ii) $2{\tan }^2{45}^{\circ }+{\cos }^2{30}^{\circ }-{\sin }^2{60}^{\circ }$

Compute step by step (digit-by-digit):

• $\tan {45}^{\circ }=1\Rightarrow {\tan }^2{45}^{\circ }=1^2=\mathrm{1<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$So $2{\tan }^2{45}^{\circ }=2\cdot 1=2.$
  

• $\cos {30}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}\Rightarrow {\cos }^2{30}^{\circ }={\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2={\displaystyle \frac{3}{4}}\mathrm{}$

• $\sin {60}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}\Rightarrow {\sin }^2{60}^{\circ }={\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2={\displaystyle \frac{3}{4}}\mathrm{}$
  

Thus ${\cos }^2{30}^{\circ }-{\sin }^2{60}^{\circ }={\displaystyle \frac{3}{4}}-{\displaystyle \frac{3}{4}}=0$

So whole expression $=2+0=2$

Answer: $2$.

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(iii) ${\displaystyle \frac{\cos {45}^{\circ }}{\sec {30}^{\circ }+\mathrm{cosec}{30}^{\circ }}}$

Compute values:

• $\cos {45}^{\circ }={\displaystyle \frac{\sqrt{2}}{2}}\mathrm{}$
  

• $\sec {30}^{\circ }={\displaystyle \frac{1}{\cos {30}^{\circ }}}={\displaystyle \frac{1}{\sqrt{3}\mathrm{/}2}}={\displaystyle \frac{2}{\sqrt{3}}}\mathrm{}$
  

• $\mathrm{cosec}{30}^{\circ }={\displaystyle \frac{1}{\sin {30}^{\circ }}}={\displaystyle \frac{1}{1\mathrm{/}2}}=2$
  

Denominator $={\displaystyle \frac{2}{\sqrt{3}}}+2={\displaystyle \frac{2+2\sqrt{3}}{\sqrt{3}}}={\displaystyle \frac{2(1+\sqrt{3})}{\sqrt{3}}}\mathrm{.}$

So

$$\frac{\cos {45}^{\circ }}{\sec {30}^{\circ }+\mathrm{cosec}{30}^{\circ }}=\frac{\sqrt{2}\mathrm{/}2}{2(1+\sqrt{3})\mathrm{/}\sqrt{3}}=\frac{\sqrt{2}}{4}\cdot \frac{\sqrt{3}}{1+\sqrt{3}}=\frac{\sqrt{6}}{4(1+\sqrt{3})}\mathrm{.}$$

Rationalize (multiply numerator & denominator by $\sqrt{3}-1$):

$$\frac{\sqrt{6}}{4(1+\sqrt{3})}\cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{6}(\sqrt{3}-1)}{4(3-1)}=\frac{\sqrt{6}(\sqrt{3}-1)}{8}\mathrm{.}$$

Answer: ${\displaystyle \frac{\sqrt{6}(\sqrt{3}-1)}{8}}$ (equivalently ${\displaystyle \frac{\sqrt{6}}{4(1+\sqrt{3})}}$

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(iv) ${\displaystyle \frac{\sin {30}^{\circ }+\tan {45}^{\circ }-\mathrm{cosec}{60}^{\circ }}{\sec {30}^{\circ }+\cos {60}^{\circ }+\cot {45}^{\circ }}}$

Compute each value:

• $\sin {30}^{\circ }=\frac{1}{2}\mathrm{}$

• $\tan {45}^{\circ }=1$

• $\mathrm{cosec}{60}^{\circ }={\displaystyle \frac{1}{\sin {60}^{\circ }}}={\displaystyle \frac{1}{\sqrt{3}\mathrm{/}2}}={\displaystyle \frac{2}{\sqrt{3}}}\mathrm{}$

So numerator $N=\frac{1}{2}+1-{\displaystyle \frac{2}{\sqrt{3}}}=\frac{3}{2}-{\displaystyle \frac{2}{\sqrt{3}}}\mathrm{}$

Denominator:

• $\sec {30}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}}\mathrm{}$

• $\cos {60}^{\circ }=\frac{1}{2}\mathrm{}$

• $\cot {45}^{\circ }=1$

So denominator $D={\displaystyle \frac{2}{\sqrt{3}}}+\frac{1}{2}+1=\frac{3}{2}+{\displaystyle \frac{2}{\sqrt{3}}}\mathrm{.}$

Thus

$$\frac{N}{D}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}\mathrm{.}$$

Multiply numerator and denominator by $2\sqrt{3}$ to clear fractions:

$$\frac{N}{D}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\mathrm{.}$$

(You can leave it in that exact form or approximate: $\approx 0.13006$.)

Answer: ${\displaystyle \frac{3\sqrt{3}-4}{3\sqrt{3}+4}}$

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(v) ${\displaystyle \frac{5{\cos }^2{60}^{\circ }+4{\sec }^2{30}^{\circ }-{\tan }^2{45}^{\circ }}{{\sin }^2{30}^{\circ }+{\cos }^2{30}^{\circ }}}$

(That is the expression as printed in the PDF.)

Compute step-by-step:

• $\cos {60}^{\circ }=\frac{1}{2}\Rightarrow {\cos }^2{60}^{\circ }=\frac{1}{4}\Rightarrow 5{\cos }^2{60}^{\circ }=5\cdot \frac{1}{4}=\frac{5}{4}\mathrm{.}$
  

• $\sec {30}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}}\Rightarrow {\sec }^2{30}^{\circ }={\displaystyle \frac{4}{3}}\Rightarrow 4{\sec }^2{30}^{\circ }=4\cdot {\displaystyle \frac{4}{3}}={\displaystyle \frac{16}{3}}\mathrm{.}$
  

• $\tan {45}^{\circ }=1\Rightarrow {\tan }^2{45}^{\circ }=1$
  

So numerator $={\displaystyle \frac{5}{4}}+{\displaystyle \frac{16}{3}}-1.$ Put over common denominator $12$:

$$\frac{15}{12}+\frac{64}{12}-\frac{12}{12}=\frac{15+64-12}{12}=\frac{67}{12}\mathrm{.}$$

Denominator:
${\sin }^2{30}^{\circ }={\left(\frac{1}{2}\right)}^2=\frac{1}{4},{\cos }^2{30}^{\circ }={\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}\mathrm{}$
So denominator $=\frac{1}{4}+\frac{3}{4}=1$

Hence whole value $={\displaystyle \frac{67}{12}}\mathrm{}$

Answer: ${\displaystyle \frac{67}{12}}\mathrm{}$

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Q2. Choose the correct option and justify your choice

(Recall: $\tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}},\tan {45}^{\circ }=1$)

(i) ${\displaystyle \frac{2{\tan }^2{30}^{\circ }}{1+{\tan }^2{30}^{\circ }}}=?$

Compute: ${\tan }^2{30}^{\circ }={\displaystyle \frac{1}{3}}\mathrm{}$
Numerator $=2\cdot {\displaystyle \frac{1}{3}}={\displaystyle \frac{2}{3}}\mathrm{}$
Denominator $=1+{\displaystyle \frac{1}{3}}={\displaystyle \frac{4}{3}}\mathrm{.}$
Quotient $={\displaystyle \frac{2\mathrm{/}3}{4\mathrm{/}3}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

${\displaystyle \frac{1}{2}}$ equals $\cos {60}^{\circ }$ and also equals $\sin {30}^{\circ }$ (Both are $1\mathrm{/}2$.)
So both options (B) and (D) numerically match; the usual textbook answer gives (B) $\cos {60}^{\circ }$(but note it is also equal to $\sin {30}^{\circ }$).

Answer: (B) $\cos {60}^{\circ }$ (value $=\frac{1}{2}$; also equals $\sin {30}^{\circ }$).

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(ii) ${\displaystyle \frac{1-{\tan }^2{45}^{\circ }}{1+{\tan }^2{45}^{\circ }}}=?$

$\tan {45}^{\circ }=1\Rightarrow {\tan }^2{45}^{\circ }=\mathrm{1. }$So numerator $=1-1=\mathrm{0<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Denominator </span>}$$=1+1=2.$ Quotient $=\mathrm{0<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

Answer: (D) $0$.

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(iii) “$\sin 2A=2\sin \mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">”\; —\; for\; which </span>}$$A$ among choices $0^{\circ },{30}^{\circ },{45}^{\circ },{60}^{\circ }$

Use identity $\sin 2A=2\sin A\cos A$. Hence we need
$2\sin A\cos A=2\sin A$. Subtract $2\sin A$: $2\sin A(\cos A-1)=0$. So either $\sin A=0$ or $\cos A=1$. Both give $A=0^{\circ }$ (in the given choice list).

Answer: (A) $0^{\circ }$.

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(iv) ${\displaystyle \frac{2{\tan }^2{30}^{\circ }}{1-{\tan }^2{30}^{\circ }}}=?$

Compute: ${\tan }^2{30}^{\circ }=\frac{1}{3}$. Numerator $=2\cdot \frac{1}{3}=\frac{2}{3}\mathrm{<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord"><span\; class="mfrac"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span></span></span></span><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Denominator </span>}$$=1-\frac{1}{3}=\frac{2}{3}\mathrm{<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Quotient </span>}$$=1$

Which option equals $1$? Among the listed options in the book, none of $\cos {60}^{\circ },\sin {60}^{\circ },\tan {60}^{\circ },\sin {30}^{\circ }$ equals $1$. So the numeric value is $1$ — if forced to choose from those printed options the closest intended match in many sources is (A) $\cos {60}^{\circ }$ is incorrect; the correct result is $1$ (so none of the four multiple-choice labels equals 1). (If the printed MCQ had different labels, pick the one equal to $1$.)

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Q3. If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)={\displaystyle \frac{1}{\sqrt{3}}}$; $0^{\circ }<A+B\le {90}^{\circ }$; $A>B$. Find $A$ and $B$.

Let $x=A+B,y=A-B\mathrm{}$ Then $\tan x=\sqrt{3}\Rightarrow x={60}^{\circ }$ (since $0^{\circ }<x\le {90}^{\circ }$).
$\tan y={\displaystyle \frac{1}{\sqrt{3}}}\Rightarrow y={30}^{\circ }$ (acute value).

Now

$$A=\frac{x+y}{2}=\frac{{60}^{\circ }+{30}^{\circ }}{2}={45}^{\circ },B=\frac{x-y}{2}=\frac{{60}^{\circ }-{30}^{\circ }}{2}={15}^{\circ }\mathrm{.}$$

Answer: $A={45}^{\circ },B={15}^{\circ }\mathrm{.}$

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Q4. State whether the following are true or false. Justify.

(i) $\sin (A+B)=\sin A+\sin B\mathrm{}$ — False.
Counterexample: let $A=B={30}^{\circ }$.

LHS= $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}\approx 0.8660$.

RHS= $\sin {30}^{\circ }+\sin {30}^{\circ }=\frac{1}{2}+\frac{1}{2}=\mathrm{1<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Not\; equal.</span>}$

(ii) The value of $\sin \theta$ increases as $\theta$ increases. — True (for $0^{\circ }\le \theta \le {90}^{\circ }$. On $[0^{\circ },{90}^{\circ }]$ sine is strictly increasing (you can see ${\displaystyle \frac{d}{d\theta }}\sin \theta =\cos \theta >0$ there, or check values from the standard table).

(iii) The value of $\cos \theta$ increases as $\theta$ increases. — False (for $0^{\circ }\le \theta \le {90}^{\circ }$, $\cos \theta$ decreases from $1$to $0$).

(iv) $\sin \theta =\cos \theta$ for all $\theta$. — False. They are equal only at specific angles (e.g. $\theta ={45}^{\circ }$ in $[0^{\circ },{90}^{\circ }]$, not for all $\theta$.

(v) $\cot A$ is not defined for $A=0^{\circ }$. — True. $\cot A={\displaystyle \frac{\cos A}{\sin A}}$ and $\sin 0^{\circ }=0$ so cotangent is undefined at $0^{\circ }$