Question

Both or questions. If you can
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Answer 1

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Answer 2

Step-by-step explanation:

• Given sec theta + tan theta = p-----------------1
• We know that (sec theta + tan theta) (sec theta – tan theta) = 1
•                        P (sec theta – tan theta) = 1
•                  Or  sec theta – tab theta = 1/p ---------------2
• Adding and subtracting we get  
• 2 sec theta = p^2 + 1 / p
• 2 tan theta = p^2 – 1 / p
• Dividing the above equations we get
• 2 tan theta / 2 sec theta = p^2 – 1 / p / p^2 + 1 / p
• So sin theta / cos theta / 1 / cos theta = p^2 – 1 / p  / p^2 + 1 / p
• Therefore sin theta =  p^2 – 1 / p^2 + 1
• If A + B = 90 degree, then prove that √tan A tan B + tan A cot B / sin A sec B – sin^2B/ cos^2 A = tan A
• We know that according to question A + B = 90 degree
•                                 So B = 90 – A
• Now given √tan A tan B + tan A cot B / sin A sec B – sin^2B/ cos^2 A
• =√tan A tan (90 – A) + tan A cot (90 – A) / sin A sec (90 – A) – sin^2 (90 – A) / cos^2 A
• = √tan A cot A + tan A. tan A / sin A cosec A – cos^2 A / cos^2 A
• = √1 + tan^2 A/ 1 – 1
• = √tan^2 A
• = tan A hence proved
• So l. h.s = r. h.s

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