Question

Prove the following$\sin \left( {{{50}^ \circ } + \theta } \right) - \cos \left( {{{40}^ \circ } - \theta } \right) + \tan {1^ \circ }\tan {10^ \circ }\tan {20^ \circ }\tan {70^ \circ }\tan {80^ \circ }\tan {89^ \circ } = 1$

Answer 1

Hint : The trigonometric function is the function that relates the ratio of the length of two sides with the angles of the right-angled triangle widely used in navigation, oceanography, the theory of periodic functions, and projectiles. Commonly used trigonometric functions are the sine, the cosine, and the tangent, whereas the cosecant, the secant, the cotangent are their reciprocal, respectively.
The value of these functions can be determined by the relation of the sides of a right-angled triangle. Trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined

Complete step-by-step answer :
In this question, identify the trigonometric identities and covert those trigonometric function in the same function and then solve the functions to prove LHS and RHS of the given equation. We try to get the same trigonometric functions everywhere so as to get the desired result.

Some of the trigonometric identities are:

$$\cos \theta = \sin \left( {90 - \theta } \right) \\\ \sin \theta = \cos \left( {90 - \theta } \right) \\\ \cot \theta = \tan \left( {90 - \theta } \right) \\\ \tan \theta = \cot \left( {90 - \theta } \right) \\\ \sec \theta = \cos \left( {90 - \theta } \right) \\\ \cos \theta = \sec \left( {90 - \theta } \right) \\\$$

The given equation
$\sin \left( {{{50}^ \circ } + \theta } \right) - \cos \left( {{{40}^ \circ } - \theta } \right) + \tan {1^ \circ }\tan {10^ \circ }\tan {20^ \circ }\tan {70^ \circ }\tan {80^ \circ }\tan {89^ \circ } = 1$
In the given equation identify the trigonometric identities and bring them together

$\sin \left( {{{50}^ \circ } + \theta } \right) - \cos \left( {{{40}^ \circ } - \theta } \right) + \tan {1^ \circ }\tan {89^ \circ }\tan {10^ \circ }\tan {80^ \circ }\tan {20^ \circ }\tan {70^ \circ } = 1$
Since${50^ \circ } = 90 - {40^ \circ }$hence we can write the function as

$\sin \left( {{{90}^ \circ } - \left( {{{40}^ \circ } - \theta } \right)} \right) - \cos \left( {{{40}^ \circ } - \theta } \right) + \tan {1^ \circ }\tan \left( {{{90}^ \circ } - {1^ \circ }} \right)\tan {10^ \circ }\tan \left( {{{90}^ \circ } - {{10}^ \circ }} \right)\tan {20^ \circ }\tan \left( {{{90}^ \circ } - {{20}^ \circ }} \right) = 1$
By using the trigonometric identities$\sin \theta = \cos \left( {90 - \theta } \right)$and$\cot \theta = \tan \left( {90 - \theta } \right)$ function can be written as

$\cos \left( {{{40}^ \circ } - \theta } \right) - \cos \left( {{{40}^ \circ } - \theta } \right) + \tan {1^ \circ }\cot {1^ \circ }\tan {10^ \circ }\cot {10^ \circ }\tan {20^ \circ }\cot {20^ \circ } = 1$
Now use reciprocal identity$\left( {\cot \theta = \dfrac{1}{{\tan \theta }}} \right)$to solve the equation further, hence we can write
$0 + \tan {1^ \circ }\dfrac{1}{{\tan {1^ \circ }}}\tan {10^ \circ }\dfrac{1}{{\tan {{10}^ \circ }}}\tan {20^ \circ }\dfrac{1}{{\tan {{20}^ \circ }}} = 1$
Now further solving the equation we get

$$0 + 1 = 1 \\\ 1 = 1 \\\$$

L.H.S=R.H.S
Hence proved

Note : Alternatively, the given trigonometric function can be reduced to a smaller expression by carrying out general algebraic and trigonometric identities. In general, this type of question can easily be solved by using trigonometric identities only.