Question

3 sec θ+5tan θ=8 then positive value of sec θ+3 tan θ is equal to:​

Answer 1

Step-by-step explanation:

5 sec A - 3tan A = 5

=> 5 √(tan² A +1) = 5 + 3 tanA

=> 25( tan² A + 1) = 25 + 9 tan² A + 30 tanA

=> 25 tan² A - 9tan² A = 30 tanA

=> 16 tan² A - 30 tanA = 0

=> 2tanA ( 8 tanA - 15 ) = 0

=> tanA = 0 or tanA = 15/8 …………. (1)

If tan A = 0 ………….. (1)

5 sec A - 3 tan A = 5

=> 5sec A -3 * 0 = 5

=> sec A = 1 ………… (1)’

If tan A = 15/8 ……… (2)

5sec A - 3* 15/8 = 5

=> 5 sec A = 5 + 45/8

=> 5 secA = 85 /8

=> sec A = 17/8 ………… (2)’

So, 5 tanA - 3 sec A = 5 * 0 - 3* 1 = -3

Or, 5*15/8 - 3*17/8 = 75/8 - 51/8 = 24/8 = 3

Answer 2

Answer: 21.96

Step-by-step explanation:

• We know,

      ∴   $sec^{2}\alpha -tan^{2}\alpha =1\\ tan\ \alpha =\sqrt{sec^{2}\alpha-1}$

• $5tan\ \alpha =8-3sec\ \alpha$

    ⇒$(5\sqrt{sec^{2}\alpha-1})^{2}=(8-3sec\ \alpha )^{2}$

    ⇒$25sec^{2}\alpha -25=64+9sec^{2}\alpha -48sec\ \alpha$

    ⇒$16sec^{2}\alpha +48sec\ \alpha=89$

    ⇒$16sec\ \alpha (sec\ \alpha +3)=89$

    ⇒$sec\ \alpha =\frac{89}{16} and sec\ \alpha =-3$

If $sec\ \alpha =\frac{89}{16}$ then we can say H=89 and B=16.

By pythogoras theorem, P=87.5

So, $tan\ \alpha =\frac{87.5}{16}$

• Now,$sec\ \alpha +3tan\ \alpha$

       ⇒ $\frac{89}{16}+\frac{262.5}{16}$

           ⇒ $\frac{89+262.5}{16}$

           ⇒ $\frac{351.5}{16}$

          ⇒ 21.96    

• Hence, the required result is 21.96.

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