त्रिकोणमिति

अभ्यास 8.2 Part 1

प्रश्न 1: निम्नलिखित के मान निकालिए:

(i) sin 60^o cos 30^o + sin 30^o cos 60^o

उत्तर:

`sin60°cos30°+sin30°cos60°`

`=sqrt3/2xxsqrt3/2+1/2xx1/2`

`=3/4+1/4=1`

(ii) 2 tan² 45^o + cos² 30^o - sin² 60^o

उत्तर:

`tan^2\45°+cos^2\30°-sin^2\60°`

`=2xx1^2+(sqrt3/2)^2-(sqrt3/2)^2`

`=2+0=2`

(iii) `(cos 45°)/(sec 30°+text(cosec) 30°)`

उत्तर:

`(cos 45°)/(sec 30°+text(cosec) 30°)`

`=(1/sqrt2)/(2/sqrt3+2)=(1/sqrt2)/((2+2sqrt3)/(sqrt3))`

`=1/sqrt2xx(sqrt3)/(2(sqrt3+1))`

`=(sqrt3)/(2sqrt2(sqrt3+1))`

`=(sqrt3)/(2sqrt2(sqrt3+1))xx(2sqrt2(sqrt3-1))/ (2sqrt2(sqrt3+1))`

`=(sqrt3xx2sqrt2(sqrt3-1))/(8(3-1))`

`=(6sqrt2-2sqrt6)/(16)`

`=(2(3sqrt2-sqrt6))/(16)`

`=(3sqrt2-sqrt6)/(8)`

(iv) `(sin30°+tan45°-text(cosec) 60°)/(Sec30°+cos60°+cot45°)`

उत्तर:

`(sin30°+tan45°-text(cosec) 60°)/(Sec30°+cos60°+cot45°)`

`=(1/2+1-2/sqrt3)/(2/sqrt2+1/2+1)`

`=(3/2-2/sqrt3)/(3/2+2/sqrt3)`

`=(3/2-2/sqrt3)/(3/2+2/sqrt3)xx(3/2-2/sqrt3)/(3/2-2/sqrt3)`

`=(9/4+4/3-2xx3/2xx2/sqrt3)/(9/4-4/3)`

`=((27+16)/(12)-2sqrt3)/((27-16)/(12))`

`=(43-24sqrt3)/(11)`

(v) `(5cos^2\60°+4sec^2\30°-tan^2\45°)/(sin^2\30°+cos^2\30°)`

उत्तर:

`(5cos^2\60°+4sec^2\30°-tan^2\45°)/(sin^2\30°+cos^2\30°)`

`=(5xx(1/2)^2+4xx(2/sqrt3)^2-1^2)/((1/2)^2+(sqrt3/2)^2)`

`=(5/4+16/3-1)/(1/4+3/4)`

`=(5/4+16/3-1)/(1)`

`=(15+64-12)/(12)=(67)/12`