Introduction to trigonometry (त्रिकोणमिति का परिचय) - Class 10th - 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°.

year 2020,2018 of 1 marks

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

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

year 2016,2010 of 1 marks

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

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

year 2014 of 1 marks

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

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

year 2012 of 1 marks

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

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

year 2019,2017,2013 of 1 marks

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

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

Q6. Evaluate {मान निकालिए}:

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

year 2015 of 1 marks

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

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

Q7. Evaluate {मान निकालिए}:

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

year 2011,2009 of 1 marks

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

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

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

year 2020 of 2 marks

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°

year 2018,2014 of 2 marks

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}:

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

year 2016,2011 of 2 marks

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}

year 2019 of 2 marks

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}

year 2017 of 2 marks

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}

year 2015 of 2 marks

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}

year 2009 of 2 marks

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}

year 2013 of 2 marks

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

year 2010 of 2 marks

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}

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

year 2012 of 2 marks

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}:
$(cosecA-cotA)^{2}=\frac{1-cosA}{1+cosA}$

year 2020,2018,2011 of 3 marks

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}:
$\frac{1-cosA}{1+cosA}=(cosecA-cotA)^{2}$

year 2014 of 3 marks

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}:
$\sqrt{\frac{1+sinA}{1-sinA}}=secA+tanA$

year 2019,2015 of 3 marks

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}:
$\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}=2secA$

year 2017,2009 of 3 marks

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}:

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

year 2016,2012 of 3 marks

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}:

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

year 2013 of 3 marks

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}:

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

year 2010 of 3 marks

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

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

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

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

$=1+cosA$

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

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

9113323460

Paytm

I hope you like it. If you like then please share it and you can also Donate to our website by my number and QR code which is given above.

Thanks.

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

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

Leave a Reply