We read your photo as:

Find the maximum and minimum values of the following, giving the values of θ between 0° and 360° for which they occur:

Mathematics

ScanSolve AI Answer

Answer

1+t1−t\frac{1+t}{1-t}

Step-by-step solution

5 steps

Ili kueleza usemi (1+sin⁡θ1−sin⁡θ)\sqrt{\left(\frac{1+\sin \theta}{1-\sin \theta}\right)} kwa kutumia t=tan⁡12θt = \tan \frac{1}{2}\theta, tutatumia mbadala wa nusu-angle.

  1. Andika upya sin⁡θ\sin \theta kwa kutumia t=tan⁡θ2t = \tan \frac{\theta}{2}.

    Tunajua kwamba sin⁡θ=2t1+t2\sin \theta = \frac{2t}{1+t^2}.

  2. Badilisha sin⁡θ\sin \theta kwenye usemi uliotolewa.

    1+sin⁡θ1−sin⁡θ=1+2t1+t21−2t1+t2\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} = \sqrt{\frac{1+\frac{2t}{1+t^2}}{1-\frac{2t}{1+t^2}}}

  3. Rahisisha namba na denominata ndani ya mzizi wa mraba.

    Namba:

    1+2t1+t2=1+t21+t2+2t1+t2=1+2t+t21+t2=(1+t)21+t21+\frac{2t}{1+t^2} = \frac{1+t^2}{1+t^2} + \frac{2t}{1+t^2} = \frac{1+2t+t^2}{1+t^2} = \frac{(1+t)^2}{1+t^2}

    Denominata:

    1−2t1+t2=1+t21+t2−2t1+t2=1−2t+t21+t2=(1−t)21+t21-\frac{2t}{1+t^2} = \frac{1+t^2}{1+t^2} - \frac{2t}{1+t^2} = \frac{1-2t+t^2}{1+t^2} = \frac{(1-t)^2}{1+t^2}

  4. Badilisha misemo iliyorahisishwa kwenye mzizi wa mraba na urahisishe zaidi.

    (1+t)21+t2(1−t)21+t2=(1+t)2(1−t)2\sqrt{\frac{\frac{(1+t)^2}{1+t^2}}{\frac{(1-t)^2}{1+t^2}}} = \sqrt{\frac{(1+t)^2}{(1-t)^2}}

  5. Chukua mzizi wa mraba.

    (1+t)2(1−t)2=∣1+t1−t∣\sqrt{\frac{(1+t)^2}{(1-t)^2}} = \left|\frac{1+t}{1-t}\right|

    Kwa kawaida, katika matatizo kama haya, tunachukua fomu chanya, tukizingatia masharti ya kikoa ambapo usemi umefafanuliwa vizuri.

    Jibu la mwisho ni 1+t1−t\frac{1+t}{1-t}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QFind the values of θ between 0° and 360° satisfying the equation 5 sin θ = −4.QFind the value of θQObtain all the values theta between 0 and 2pi such that = (1)/(sqrt(2)).QSolve the equation sin(5/2) θ = 1/2 for 0° ≤ θ ≤ 180°.
Browse by topic:Calculus
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp