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Question: If \[\sin \theta + \cos \theta = \sqrt 2 \cos ({90^ \circ } - \theta )\] then prove that, \[\tan \th...

If sin⁡θ+cos⁡θ=2cos⁡(90∘−θ)\sin \theta + \cos \theta = \sqrt 2 \cos ({90^ \circ } - \theta ) then prove that, tan⁡θ=2+1\tan \theta = \sqrt 2 + 1

Explanation

Solution

We will solve the equation by using complementary angle theorem. Then we will interchange the side of trigonometric ratios which will be useful to arrive at the final answer. Some formulae used are being stated below.
Used formula: cos⁡(90∘−θ)=sin⁡θ\cos ({90^ \circ } - \theta ) = \sin \theta
a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b)

_Complete step-by-step answer: _
It is given that, sin⁡θ+cos⁡θ=2cos⁡(90∘−θ)\sin \theta + \cos \theta = \sqrt 2 \cos ({90^ \circ } - \theta ).
We have to prove that,tan⁡θ=2+1\tan \theta = \sqrt 2 + 1.
Consider the given equation as equation number (1).
sin⁡θ+cos⁡θ=2cos⁡(90∘−θ)\sin \theta + \cos \theta = \sqrt 2 \cos ({90^ \circ } - \theta )… (1)
We know that, by the relation between complementary angles stated by complementary angle theorem we get,
cos⁡(90∘−θ)=sin⁡θ\cos ({90^ \circ } - \theta ) = \sin \theta
Let us use the above expression in equation (1), then we get,
sin⁡θ+cos⁡θ=2sin⁡θ\sin \theta + \cos \theta = \sqrt 2 \sin \theta
Let us dividesin⁡θ\sin \theta in both the sides of the above equation we get,
sin⁡θ+cos⁡θsin⁡θ=2sin⁡θsin⁡θ\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta }} = \dfrac{{\sqrt 2 \sin \theta }}{{\sin \theta }}
Let us simplify the above equation by dividing it in every term, we get,
1+cot⁡θ=21 + \cot \theta = \sqrt 2
Let us rewrite the above equation as follows then we get,
cot⁡θ=2−1\cot \theta = \sqrt 2 - 1
Again, we know that, tan⁡θ=1cot⁡θ\tan \theta = \dfrac{1}{{\cot \theta }} and vice versa also holds and here we are going to use the vice versa part to prove the required result.
Now using the above equation let us rewrite the above found equation as follows,
tan⁡θ=12−1\tan \theta = \dfrac{1}{{\sqrt 2 - 1}}
Let us again simplify the above equation by multiplying and dividing the right hand side with 2+1\sqrt 2 + 1 , then we get,
tan⁡θ=2+1(2−1)(2+1)\tan \theta = \dfrac{{\sqrt 2 + 1}}{{(\sqrt 2 - 1)(\sqrt 2 + 1)}}
The denominator is of the form (a+b)(a−b)(a + b)(a - b) so using the formula given in the hint we get,
tan⁡θ=2+12−1\tan \theta = \dfrac{{\sqrt 2 + 1}}{{2 - 1}}
Finally we have arrived at the following equation,
tan⁡θ=2+1\tan \theta = \sqrt 2 + 1

Hence we have proved that tan⁡θ=2+1\tan \theta = \sqrt 2 + 1 .

Note:
We know that the sum of all the angles of a triangle is 180∘.{180^ \circ }. For a right-angle triangle, one of the angles is 90∘{90^ \circ } and the sum of the other two angles is 90∘{90^ \circ }. Then the complementary angle theorem states that,
sin⁡(90∘−θ)=cos⁡θ\sin ({90^ \circ } - \theta ) = \cos \theta and cos⁡(90∘−θ)=sin⁡θ\cos ({90^ \circ } - \theta ) = \sin \theta .