Question

Help please. Maths trigonometry.
Q. Of Triangle

Answer 1

In ΔABC, AB = x units   AC = 7 units
     BC = √(49 - x²) =√[(7+x)(7-x)]  units  by  Pythagoras thm.
     ∠A+∠C =  90°     => Cos A = Sin C     and  Cot A = Tan C

Tan C = Cot A = AB/BC = x/√[(7+x)(7-x)]
Cos A = Sin C = x/7

LHS = [ √(7-x) + √(7+x) ] * Tan C  + Cos A (-14 +21) + √(49+x²) * Cos 90°
         = [√(7-x) + √(7+x)] * x/√[49-x²)  +  7 * x/7 + √(49+x²) * 0
        = x [ √(7-x)  + √(7+x)] / √(49-x²)  +  x
If required,  Rationalize the denominator in the first term by multiplying Nr  and Dr by √(49-x²).

looks like we cant simplify this further.
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ΔPQR:   
PQ : QR : PR = 8 : 15 : 17
check that  PR² = PQ² + QR²
So it is a right angle triangle , they make a Pythagorean triplet.  As PR is largest, it is Hypotenuse.  ∠Q = 90°.             ∠P + ∠R = 90°
Draw the triangle with this description.

Sin R = Cos P = PQ/PR = 8/17
Sin P = Cos R = QR/PR = 15/17

LHS = 0 as  Cos P cosR = sin P sin R. 

Answer 2

Step-by-step explanation:

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