Question

If tan 35°is x than
tan215°- tan125° ÷ tan235° + tan325°

Answer 1

Answer:

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Explanation:

Given, tan 35° = k

= (tan 145° - tan 125°)/(1+tan 145°.tan 125°)

= {tan (180-35) - tan (90+35)}/{1+tan (180-35).tan (90-35)}

= {-tan 35° - (-cot 35°)}/{1+(-tan 35°)(-cot 35°)

= {-k-(-1/k)}/{1+(-k)(-1/k)}

= (-k²+1/k) ÷ (1+1)

= -k²+1/2k

= (1-k²)/2k

Answer 2

Therefore the value of (tan 215° - tan 125°) ÷ (tan 235° + tan325°) comes out to be equal to (1 - x²)/(1+x²).

Given:

tan 35° = x

To Find:

The value of (tan 215° - tan 125°) ÷ (tan 235° + tan325°)

Solution:

∵ x = tan 35°

∴ cot 35° = 1/x

• All trigonometric functions are positive in the first quadrant.
• The sin and cosec functions are positive in the second quadrant.
• The tan and cot functions are positive in the third quadrant.
• The cos and sec functions are positive in the fourth quadrant.
• sin, cos, and tan change to cosec, sec, and cot respectively at odd multiples of π/2.
• cosec, sec, and cot change to sin, cos, and tan respectively at odd multiples of π/2.
• No changes occur at even multiples of π/2.

Formulas to be used:

(1.) tan (90° + α) = -cot α                               (2.) tan (180° - α) = -tan α

(3.) tan (180° + α) = tan α                              (4.) tan (270° - α) = cot α  

(5.) tan (270° + α) = -cot α                            (6.) tan (360° - α) = -tanα

Procedure:

                                $=\frac{tan215^0-tan125^0}{tan235^0+tan325^0} \\\\=\frac{tan(180^0+35^0)+tan(90^0+35^0)}{tan(270^0-35^0)+tan(360^0-35^0)} \\\\=\frac{tan35^0-cot35^0}{-cot35^0-tan35^0} \\\\=\frac{x-(\frac{1}{x} )}{-(\frac{1}{x} )-x} =\frac{x^2-1}{-1-x^2} =\frac{1-x^2}{1+x^2}$

Therefore the value of (tan 215° - tan 125°) ÷ (tan 235° + tan325°) comes out to be equal to (1 - x²)/(1+x²).

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