Half Angle Formula Required?

Geometry Level 3

For θ ∈ ( 0 , π 2 ) \theta\in\left(0,\frac{\pi}{2}\right) , which of the following is equivalent to the expression given below? tan ⁡ θ + sec ⁡ θ − 1 tan ⁡ θ − sec ⁡ θ + 1 \frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}

This problem is part of the set Trigonometry .

1 − sin ⁡ θ cos ⁡ θ \frac{1-\sin\theta}{\cos\theta} 1 + cos ⁡ θ sin ⁡ θ \frac{1+\cos\theta}{\sin\theta} 1 + sin ⁡ θ cos ⁡ θ \frac{1+\sin\theta}{\cos\theta} cos ⁡ θ 1 + sin ⁡ θ \frac{\cos\theta}{1+\sin\theta}

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Omkar Kulkarni by Omkar Kulkarni , 22, India

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5 solutions

Roman Frago
Feb 10, 2015

12 Helpful 0 Interesting 0 Brilliant 0 Confused

your method is same as mine

Ramez Hindi - 6 years, 4 months ago

how to get that 1st = ?

宥 神 - 6 years, 4 months ago

Multiply the numerator and denominator of the given expression by tan ⁡ ( θ ) − sec ⁡ ( θ ) − 1. \tan(\theta) - \sec(\theta) - 1. For the numerator we then have

( ( t a n ( θ ) + sec ⁡ ( θ ) ) − 1 ) ( ( tan ⁡ ( θ ) − sec ⁡ ( θ ) ) − 1 ) = ((tan(\theta) + \sec(\theta)) - 1)((\tan(\theta) - \sec(\theta)) - 1) =

tan ⁡ 2 ( θ ) − sec ⁡ 2 ( θ ) − 2 tan ⁡ ( θ ) + 1 = − 2 tan ⁡ ( θ ) \tan^{2}(\theta) - \sec^{2}(\theta) - 2\tan(\theta) + 1 = -2\tan(\theta)

since 1 + tan ⁡ 2 ( θ ) = sec ⁡ 2 ( θ ) . 1 + \tan^{2}(\theta) = \sec^{2}(\theta). For the denominator we then have

( t a n ( θ ) − sec ⁡ ( θ ) ) + 1 ) ( ( tan ⁡ ( θ ) − sec ⁡ ( θ ) ) − 1 ) = (tan(\theta) - \sec(\theta)) + 1)((\tan(\theta) - \sec(\theta)) - 1) =

tan ⁡ 2 ( θ ) + sec ⁡ 2 ( θ ) − 2 tan ⁡ ( θ ) sec ⁡ ( θ ) − 1 = \tan^{2}(\theta) + \sec^{2}(\theta) - 2\tan(\theta)\sec(\theta) - 1 =

2 tan ⁡ 2 ( θ ) − 2 tan ⁡ ( θ ) sec ⁡ ( θ ) = 2 tan ⁡ ( θ ) ( tan ⁡ ( θ ) − sec ⁡ ( θ ) ) 2\tan^{2}(\theta) - 2\tan(\theta)\sec(\theta) = 2\tan(\theta)(\tan(\theta) - \sec(\theta)) .

So the given expression is transformed to

− 2 tan ⁡ ( θ ) 2 tan ⁡ ( θ ) ( tan ⁡ ( θ ) − sec ⁡ ( θ ) ) = 1 sec ⁡ ( θ ) − tan ⁡ ( θ ) = cos ⁡ ( θ ) 1 − sin ⁡ ( θ ) = 1 + sin ⁡ ( θ ) cos ⁡ ( θ ) \dfrac{-2\tan(\theta)}{2\tan(\theta)(\tan(\theta) - \sec(\theta))} =\dfrac{1}{\sec(\theta) - \tan(\theta)} = \dfrac{\cos(\theta)}{1 - \sin(\theta)} = \dfrac{1 + \sin(\theta)}{\cos(\theta)} ,

where the last step was achieved by multiplying top and bottom by ( 1 + sin ⁡ ( θ ) ) (1 + \sin(\theta)) .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

U HV done the hard way..Take theta as A --> tan A + sec A -(sec^2-tan^2)/tan A-sec A+1.....since..{1+tan^2=sec^2}.

--> tanA + secA -(secA+tanA)(secA-tanA)/tan A-sec A+1

--> (tanA+secA)(1-secA+tanA)/tan A-sec A+1

cut the common brackets out..

we are left with

=tanA+secA

=sinA/cosA + 1/cosA

sinA+1/cosA

Arnab Sinha - 6 years, 4 months ago

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Yes I also did your way , but selected the answer as mirror image of it.

U Z - 6 years, 4 months ago

Easiest way.. I did it same way..

Bhupendra Jangir - 6 years, 2 months ago
Aaaaa Bbbbb
Feb 11, 2015

s i n ( θ ) + 1 − c o s ( θ ) s i n ( θ ) − 1 + c o s ( θ ) \frac{sin(\theta)+1-cos(\theta)}{sin(\theta)-1+cos(\theta)} = 2 s i n ( θ 2 ) c o s ( θ 2 ) + 1 − 1 + 2 s i n 2 ( θ 2 ) 2 s i n ( θ 2 ) c o s ( θ 2 ) − 1 + 1 − 2 s i n 2 ( θ 2 ) =\frac{2sin(\frac{\theta}{2})cos(\frac{\theta}{2})+1-1+2sin^{2}(\frac{\theta}{2})}{2sin(\frac{\theta}{2})cos(\frac{\theta}{2})-1+1-2sin^{2}(\frac{\theta}{2})} = s i n ( θ 2 ) c o s ( θ 2 ) + s i n 2 ( θ 2 ) s i n ( θ 2 ) c o s ( θ 2 ) − s i n 2 ( θ 2 ) =\frac{sin(\frac{\theta}{2})cos(\frac{\theta}{2})+sin^{2}(\frac{\theta}{2})}{sin(\frac{\theta}{2})cos(\frac{\theta}{2})-sin^{2}(\frac{\theta}{2})} = c o s ( θ 2 ) + s i n ( θ 2 ) c o s ( θ 2 ) − s i n ( θ 2 ) =\frac{cos(\frac{\theta}{2})+sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})-sin(\frac{\theta}{2})} = 1 + s i n ( θ ) c o s ( θ ) =\boxed{\frac{1+sin(\theta)}{cos(\theta)}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

how did you achieve the last step

John Doe - 6 years, 3 months ago

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It can be by multiplying numerator and denominator with its numerator ...

aaaaaa bbbbbb - 6 years, 3 months ago
Afreen Sheikh
Apr 14, 2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ramez Hindi
Feb 10, 2015

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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