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osif cose -2 where x lies in lird quadrant then tinhd the value ot sinco ,thAn2Q6. If tanx +sinxam, tanx-sinx-n, to show that:-- 2-n-dymn.Qz 1f A B, Cand D are the angle of a cyclic quadrilateral then prove that cosA +cosBtors +costcind the maximum and minimum 'slueyi sinxtcos. Ans-V2-2u .T trreeengtesof A, B and Care iaA·P, then prove t -cotBe r-s 15-Q10.To prove that the following identiticssAHB 4-Bsin A +sin Bsin A sin Biv)sin100 sin30 osinsoo sin70%(v)sin200 sinano sin60°sin80%(vi) Sec84+(ix) Sin5xsSsinx-20sin3x +16 sínsx(xi) Cos2x 2sin'y +4cos(x+y) sinxsiny-cos2(x+y) (xii)v3cosec200-sec 20.4(kxili) cosAcos B-2 cos Acos B cos(A B sin AB)iv cos A.cos 2.A.cos 2Q11. Find the principal solution of the following functionsvi cos 1s1+cos(1cos16tan 5 tan 34CoS2Aco4Atai5A tan3 Asec 4 Atan 24never lies between 1/3 and 3fan x2 A cos2 A cos 2 A2 sin 47π 11πsee2 Anstanxe Ans |
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Answer» I dont have enough time i had to submit assignment with in 2 hr Plz hlp me Are working. For it or not Ok.. I'll post the answer Suno serialwise karo 5)since,sin60=√ 3/2 = √ 3/2( sin20sin40sin80) =√ 3/2( sin20sin80sin40) =√ 3/4 [(2sin20sin40)sin80] on applying [cos(A-B)-cos(A+B) = 2sinAsinB] we get, = √ 3/4 (cos20-cos60)sin80 [since,cos(-a)=cosa] = √ 3/4(cos20sin80-cos60sin80) = √ 3/8(2sin80cos20-sin80) = √ 3/8(sin100+sin60-sin80) = √ 3/8( √ 3/2+sin100-sin80 ) = √ 3/8( √ 3/2+sin(180-80)-sin80 ) = √ 3/8( √ 3/2+sin80-sin80 ) [since,sin(180-a)=sina] = √ 3/8( √ 3/2) = 3/16 Saare karo solve Solve the whole questions |
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