if 7 sin^2 theta + 3 cos ^2 theta =4 .show tan theta =1/root3

7sin2ϴ + 3cos2ϴ = 4

4sin2 ϴ + 3sin2 ϴ+ 3cos2 ϴ = 4

4sin2 ϴ +3 (sin2 ϴ + cos2 ϴ) = 4

4sin2 ϴ + 3 = 4

4sin2 ϴ = 1 

sin2 ϴ =1/4 

sin ϴ = ½

sin ϴ =perp./hypo. = ½

So, perp. = 1k  ; hypo. = 2k

(Hypo)2  = (base)2 + (perp.)2

k2 = (base)2 + (2k)2

(base)2 = 4k2 – k2

base =  root3 k

tan ϴ = perp./base = k/root3 k

so, tan ϴ =1/root3

Hope it helps !!!!!!!!!!!!!!!!!!!!!!!!!

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i know but i cannot explain
 
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thank you!!
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Thanks
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See the photo it may help but u may not understand my handwriting

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Pheku answer of this...haha...lol...wrong answer
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That's is answer bhaiyo aur bheno

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It will help u.

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7sin2ϴ + 3cos2ϴ = 4 4sin2 ϴ + 3sin2 ϴ+ 3cos2 ϴ = 4 4sin2 ϴ +3 (sin2 ϴ + cos2 ϴ) = 4 4sin2 ϴ + 3 = 4 4sin2 ϴ = 1 sin2 ϴ =1/4 sin ϴ = ½ sin ϴ =perp./hypo. = ½ So, perp. = 1k ; hypo. = 2k (Hypo)2 = (base)2 + (perp.)2 k2 = (base)2 + (2k)2 (base)2 = 4k2 – k2 base =  root3 k tan ϴ = perp./base = k/root3 k so, tan ϴ =1/root3 Hope it helps !!!!!!!!!!!!!!!!!!!!!!!!!
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7sin2ϴ + 3cos2ϴ = 4 4sin2 ϴ + 3sin2 ϴ+ 3cos2 ϴ = 4 4sin2 ϴ +3 (sin2 ϴ + cos2 ϴ) = 4 4sin2 ϴ + 3 = 4 4sin2 ϴ = 1 sin2 ϴ =1/4 sin ϴ = ½ sin ϴ =perp./hypo. = ½ So, perp. = 1k ; hypo. = 2k (Hypo)2 = (base)2 + (perp.)2 k2 = (base)2 + (2k)2 (base)2 = 4k2 – k2 base =  root3 k tan ϴ = perp./base = k/root3 k so, tan ϴ =1/root3 Hope it helps !!!!!!!!!!!!!!!!!!!!!!
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