Introduction to trigonometry (त्रिकोणमिति का परिचय) – STUDENTHAPPY

Introduction to trigonometry (त्रिकोणमिति का परिचय)

Get “Introduction to trigonometry” chapter’s previous years questions from 2009 to 2020 of JAC board.

Q1. Evaluate {मान निकालिए}:
Cosec31°- sec59°.

[su_spoiler title=”year 2020,2018 of 1 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. Cosec31°- sec59°
= Cosec(90°- 59°) – sec59°
= Sec59°- sec59°
= 0

Q2. Evaluate {मान निकालिए}:
cos48°- sin42°

[su_spoiler title=”year 2016,2010 of 1 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. cos48°- sin42°
= cos(90°- 42°) – sin42°
= sin42°- sin42°
= 0

Q3. Evaluate {मान निकालिए}:
tan65°- cot25°

[su_spoiler title=”year 2014 of 1 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. tan65°- cot25°
= tan(90°- 25°) – cot25°
= cot25°- cot25°
= 0

Q4. Evaluate {मान निकालिए}:
sin23°- cos67°

[su_spoiler title=”year 2012 of 1 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. sin23°- cos67°
= sin(90°- 67) – cos67°
= cos67°- cos67°
= 0

Q5. Evaluate {मान निकालिए}:
$\frac{sin18\degree}{cos72\degree}$

[su_spoiler title=”year 2019,2017,2013 of 1 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $\frac{sin18\degree}{cos72\degree}$

= $\frac{sin(90\degree-72\degree)}{cos72\degree}=\frac{cos72\degree}{cos72\degree}=1$

Q6. Evaluate {मान निकालिए}:

$\frac{sin47\degree}{cos43\degree}$

[su_spoiler title=”year 2015 of 1 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $\frac{sin47\degree}{cos43\degree}$

$=\frac{sin(90\degree-43\degree)}{cos43\degree}=\frac{cos43\degree}{cos43\degree}=1$

Q7. Evaluate {मान निकालिए}:

$\frac{tan26\degree}{cot64\degree}$

[su_spoiler title=”year 2011,2009 of 1 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $\frac{tan26\degree}{cot64\degree}$

$=\frac{tan(90\degree-64\degree)}{cot64\degree}=\frac{cot64\degree}{cot64\degree}=1$

Q8. Evaluate {मान निकालिए}:
$5cos^{2}60\degree+4sec^{2}30\degree-tan^{2}45\degree$

[su_spoiler title=”year 2020 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $5cos^{2}60\degree+4sec^{2}30\degree-tan^{2}45\degree$

$=5\times\left ( \frac{1}{2} \right )^{2}+4\times\left ( \frac{2}{\sqrt{3}} \right )^{2}-1^{2}$

$=\frac{5}{4}+\frac{16}{3}-1$

$=\frac{15+64-12}{12}=\frac{79-12}{12}=\frac{67}{12}$

Q9. Evaluate {मान निकालिए}:
sin60°.cos30°+sin30°.cos60°

[su_spoiler title=”year 2018,2014 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. sin60°.cos30°+sin30°.cos60°

$=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}+\frac{1}{2}\times\frac{1}{2}$

$=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$

Q10. Evaluate {मान निकालिए}:

$2tan^{2}45\degree+cos^{2}30\degree-sin^{2}60\degree$

[su_spoiler title=”year 2016,2011 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $2tan^{2}45\degree+cos^{2}30\degree-sin^{2}60\degree$

$=2\times1^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}-\left ( \frac{\sqrt{3}}{2} \right )^{2}$

$=2+\frac{3}{4}-\frac{3}{4}=2$

Q11. If $cotA=\frac{8}{15}$ then find the value of sinA.

{यदि $cotA=\frac{8}{15}$ तो sinA का मान ज्ञात कीजिए}

[su_spoiler title=”year 2019 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $cotA=\frac{8}{15}=\frac{b}{p}$
∴ b=8, p=15

∴ $h=\sqrt{p^{2}+b^{2}}=\sqrt{15^{2}+8^{2}}$
$=\sqrt{225+64}=17$

∴ $sinA=\frac{p}{h}=\frac{15}{17}$

Q12. If $tanA=\frac{3}{4}$ then find the value of sinA, cosA

{यदि $tanA=\frac{3}{4}$ तो sinA , cosA का मान ज्ञात कीजिए}

[su_spoiler title=”year 2017 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $tanA=\frac{3}{4}=\frac{p}{b}$
∴ p=3, b=4

∴ $h=\sqrt{p^{2}+b^{2}}=\sqrt{3^{2}+4^{2}}$
$= \sqrt{9+16}=5$

∴ $sinA=\frac{p}{h}=\frac{3}{5}$ ,

$cosA=\frac{b}{h}=\frac{4}{5}$

Q13. If $tanA=\frac{3}{4}$ then find the value of cosA, cosecA

{यदि $tanA=\frac{3}{4}$ तो cosA, cosecA का मान ज्ञात कीजिए}

[su_spoiler title=”year 2015 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $tanA=\frac{3}{4}=\frac{p}{b}$
∴ p=3, b=4

∴ $h=\sqrt{p^{2}+b^{2}}=\sqrt{3^{2}+4^{2}}$
$= \sqrt{9+16}=5$

∴ $cosA=\frac{b}{h}=\frac{4}{5}$ ,

$cosecA=\frac{h}{p}=\frac{5}{3}$

Q14. If $tanA=\frac{4}{3}$ then find the value of sinA, secA

{यदि $tanA=\frac{4}{3}$ तो sinA , secA का मान ज्ञात कीजिए}

[su_spoiler title=”year 2009 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $tanA=\frac{4}{3}=\frac{p}{b}$
∴ p=4, b=3

∴ $h=\sqrt{p^{2}+b^{2}}=\sqrt{4^{2}+3^{2}}$
$=\sqrt{16+9}=5$

∴ $sinA=\frac{p}{h}=\frac{4}{5} ,$

$secA=\frac{h}{b}=\frac{5}{3}$

Q15. If $sinA=\frac{3}{4}$ then find the value of cosA

{यदि $sinA=\frac{3}{4}$ तो cosA का मान ज्ञात कीजिए}

[su_spoiler title=”year 2013 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $sinA=\frac{3}{4}=\frac{p}{h}$
∴ p=3 , h=4

∴ $b=\sqrt{h^{2}-p^{2}}=\sqrt{4^{2}-3^{2}}=\sqrt{7}$

∴ $cosA=\frac{b}{h}=\frac{\sqrt{7}}{4}$

Q16. Evaluate {मान निकालिए}:
$\frac{2tan30\degree}{1-tan^{2}30\degree}$

[su_spoiler title=”year 2010 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $\frac{2tan30\degree}{1-tan^{2}30\degree}$

$=\frac{2\times\frac{1}{\sqrt{3}}}{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}}$ $=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$

$=\frac{2}{\sqrt{3}}\times\frac{3}{2}=\sqrt{3}$

Q17. Evaluate. {मान निकालिए}

$\frac{sin30\degree+tan45\degree-cosec60\degree}{sec30\degree+cos60\degree+cot45\degree}$

[su_spoiler title=”year 2012 of 2 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $\frac{sin30\degree+tan45\degree-cosec60\degree}{sec30\degree+cos60\degree+cot45\degree}$

$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}$

$=\frac{\frac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\frac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}$

$=\frac{3\sqrt{3}-4}{4+3\sqrt{3}}$

$=\frac{3\sqrt{3}-4}{4+3\sqrt{3}}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}$

$=\frac{12\sqrt{3}-27-16+12\sqrt{3}}{16-27}$

$=\frac{24\sqrt{3}-43}{-11}$

$=\frac{43-24\sqrt{3}}{11}$

Q18. Prove that {सिद्ध कीजिए}:
$(cosecA-cotA)^{2}=\frac{1-cosA}{1+cosA}$

[su_spoiler title=”year 2020,2018,2011 of 3 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. L.H.S = $(cosecA-cotA)^{2}$

$=\left ( \frac{1}{sinA}-\frac{cosA}{sinA} \right )^{2}$

$=\left ( \frac{1-cosA}{sinA} \right )^{2}$

$=\frac{(1-cosA)^{2}}{sin^{2}A}$

$=\frac{(1-cosA)^{2}}{1-cos^{2}A}$

$=\frac{(1-cosA)(1-cos)}{(1-cosA)(1+cosA)}$

$=\frac{(1-cos)}{(1+cosA)}$ $=R.H.S$

Q19. Prove that {सिद्ध कीजिए}:
$\frac{1-cosA}{1+cosA}=(cosecA-cotA)^{2}$

[su_spoiler title=”year 2014 of 3 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. L.H.S $=\frac{1-cosA}{1+cosA}$

Multiply numerator and denominator by (1-cosA)
{मीटर और विभाजक को (1-cosA) से गुणा करने पर}

$=\frac{1-cosA}{1+cosA}\times\frac{1-cosA}{1-cosA}$

$=\frac{(1-cosA)^{2}}{1-cos^{2}A}$

$=\frac{(1-cosA)^{2}}{sin^{2}A}$

$=\left ( \frac{1-cosA}{sinA} \right )^{2}$

$=\left ( \frac{1}{sinA}-\frac{cosA}{sinA} \right )^{2}$

$=\left ( cosecA-cotA \right )^{2}$ $=R.H.S$

Q20. Prove that {सिद्ध कीजिए}:
$\sqrt{\frac{1+sinA}{1-sinA}}=secA+tanA$

[su_spoiler title=”year 2019,2015 of 3 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $L.H.S=\sqrt{\frac{1+sinA}{1-sinA}}$

Multiply numerator and denominator by (1+sinA)
{मीटर और विभाजक को (1+sinA) से गुणा करने पर}

$=\sqrt{\frac{1+sinA}{1-sinA}\times\frac{1+sinA}{1+sinA}}$

$=\sqrt{\frac{(1+sinA)^{2}}{1^{2}-sin^{2}A}}$

$=\sqrt{\frac{(1+sinA)^{2}}{cos^{2}A}}$

$=\sqrt{\left ( \frac{1+sinA}{cosA} \right )^{2}}$

$=\frac{(1+sinA)}{cosA}$

$=\frac{1}{cosA}+\frac{sinA}{cosA}$

$=secA+tanA$ $=R.H.S$

Q21. Prove that {सिद्ध कीजिए}:
$\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}=2secA$

[su_spoiler title=”year 2017,2009 of 3 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $L.H.S=\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}$

$=\frac{cos^{2}A+(1+sinA)^{2}}{(1+sinA)cosA}$

$=\frac{cos^{2}A+1+sin^{2}A+2sinA}{(1+sinA)cosA}$

$=\frac{sin^{2}A+cos^{2}A+1+2sinA}{(1+sinA)cosA}$

$=\frac{1+1+2sinA}{(1+sinA)cosA}$

$=\frac{2+2sinA}{(1+sinA)cosA}$

$=\frac{2(1+sinA)}{(1+sinA)cosA}$

$=2\times \frac{1}{cosA}$

$=2secA$ $=R.H.S$

Q22. Prove that {सिद्ध कीजिए}:

$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=1+secA.cosecA$

[su_spoiler title=”year 2016,2012 of 3 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $L.H.S=\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$

$=\frac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}+\frac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}$

$=\frac{\frac{sinA}{cosA}}{\frac{sinA-cosA}{sinA}}+\frac{\frac{cosA}{sinA}}{\frac{cosA-sinA}{cosA}}$

$=\frac{{sin^{2}A}}{cosA({sinA-cosA})}+\frac{{cos^{2}A}}{sinA(cosA-sinA)}$

$=\frac{{sin^{2}A}}{cosA({sinA-cosA})}-\frac{{cos^{2}A}}{sinA(sinA-cosA)}$

$=\frac{sin^{3}A-cos^3A}{sinAcosA(sinA-cosA)}$

$Using\, the\, formula\, [a^3-b^3=(a-b)(a^2+b^2+ab)]$
{सूत्र [ $a^3-b^3=(a-b)(a^2+b^2+ab)$ ]से}

$=\frac{(sinA-cosA)(sin^2A+cos^2A+sinAcosA)}{sinAcosA(sinA-cosA)}$

$=\frac{sin^{2}A+cos^{2}A+sinA.cosA}{sinA.cosA}$

$=\frac{1+sinAcosA}{sinA.cosA}$ $[\because sin^{2}A+cos^{2}A=1]$

$=\frac{1}{sinA.cosA}+\frac{sinA.cosA}{sinA.cosA}$

$=cosecA.secA+1$

$=1+cosecA.secA=R.H.S$ $(proved)$

OR

$L.H.S=\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$

$=\frac{tanA}{1-\frac{1}{tanA}}+\frac{\frac{1}{tanA}}{1-tanA}$

$=\frac{tanA}{\frac{tanA-1}{tanA}}+\frac{1}{tanA(1-tanA)}$

$=\frac{tan^{2}A}{tanA-1}+\frac{1}{tanA(1-tanA)}$

$=\frac{tan^{2}A}{tanA-1}-\frac{1}{tanA(tanA-1)}$

$=\frac{tan^{3}A-1}{tanA(tanA-1)}$

$Using\, the\, formula\,[a^3-b^3=(a-b)(a^2+ab+b^2)]$
{सूत्र [ $a^3-b^3=(a-b)(a^2+ab+b^2)$ ] से}

$=\frac{(tanA-1)(tan^{2}A+tanA+1)}{tanA(tanA-1)}$

$=\frac{(tan^{2}A+tanA+1)}{tanA}$

$=\frac{tan^{2}A}{tanA}+\frac{tanA}{tanA}+\frac{1}{tanA}$

$={tanA}+1+cotA$

$=1+{tanA}+cotA$

$=1+\left ( \frac{sinA}{cosA}+\frac{cosA}{sinA}\right )$

$=1+\left ( \frac{sin^{2}A+cos^{2}A}{sinA.cosA}\right )$

$=1+\left ( \frac{1}{sinA.cosA}\right )$

$=1+\frac{1}{sinA}.\frac{1}{cosA}$

$=1+cosecA.secA=R.H.S$ $(proved)$

Q23. Prove that {सिद्ध कीजिए}:

$(secA+tanA)(1-sinA)=cosA$

[su_spoiler title=”year 2013 of 3 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $L.H.S=(secA+tanA)(1-sinA)$

$=\left ( \frac{1}{cosA}+\frac{sinA}{cosA} \right )\left ( 1-sinA\right )$

$=\left ( \frac{1+sinA}{cosA} \right )(1-sinA)$

$=\left ( \frac{(1+sinA)(1-sinA)}{cosA} \right )$

$=\frac{(1-sin^{2}A)}{cosA}$

$=\frac{(cos^{2}A)}{cosA}$

$={cosA}=R.H.S$

Q24. Prove that {सिद्ध कीजिए}:

$\frac{1+secA}{secA}=\frac{sin^{2}A}{1-cosA}$

[su_spoiler title=”year 2010 of 3 marks” style=”simple” anchor_in_url=”yes”][/su_spoiler]

Ans. $L.H.S=\frac{1+secA}{secA}$

$=\frac{1+\frac{1}{cosA}}{\frac{1}{cosA}}$

$=\frac{\frac{cosA+1}{cosA}}{\frac{1}{cosA}}$

$=cosA+1$

$=1+cosA$

Multiply numerator and denominator by 1-cosA
{मीटर और विभाजक को (1-cosA) से गुणा करने पर}

$=(1+cosA)\times\frac{1-cosA}{1-cosA}$

$=\frac{1-cos^{2}A}{1-cosA}$

$=\frac{sin^{2}A}{1-cosA}=R.H.S$ $(proved)$

OR

$R.H.S=\frac{sin^{2}A}{1-cosA}$

$=\frac{1-cos^{2}A}{1-cosA}$

$=\frac{(1-cosA)(1+cosA)}{1-cosA}$

$=1+cosA$

$=1+\frac{1}{secA}$

$=\frac{secA+1}{secA}=L.H.S$ $(proved)$

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Next Chapter $\rightarrow$ Some applications of trigonometry (त्रिकोणमिति के कुछ अनुप्रयोग).

Previous Chapter $\rightarrow$ Coordinate geometry (निर्देशांक ज्यामिति)

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