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Some Applications Of Trigonometry

Question
CBSEENMA10008273
Wired Faculty App

If 7 sin2θ + 3cos2θ = 4, show that tanθ space equals space fraction numerator 1 over denominator square root of 3 end fraction.

Solution

We have,
  7 space sin squared straight theta plus 3 space cos squared straight theta space equals space 4
rightwards double arrow space space 4 space sin squared straight theta plus 3 space sin squared straight theta plus 3 cos squared straight theta space equals space 4 rightwards double arrow space space 4 space sin squared straight theta space plus space 3 space left parenthesis sin squared straight theta space plus space cos squared straight theta right parenthesis space equals space 4 rightwards double arrow space 4 space sin squared straight theta space plus space 3 cross times 1 space equals space 4 rightwards double arrow space 4 space sin squared straight theta space equals space 4 space minus 3 rightwards double arrow space 4 space sin squared straight theta space equals space 1 rightwards double arrow space space sin squared straight theta space equals space 1 fourth
 And comma space cos squared straight theta space equals space 1 minus sin squared straight theta. space space space space space space space space space space space equals space 1 minus 1 fourth equals 3 over 4 Now comma space fraction numerator sin squared straight theta over denominator cos squared straight theta end fraction equals 1 third rightwards double arrow space tan squared straight theta space equals space 1 third rightwards double arrow space tanθ space equals space fraction numerator 1 over denominator square root of 3 end fraction

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