Question

Evaluate:
Sin²34 + sin²56 + 2 tan18 tan72 - cot²30

Answer 1

sin²34°+ sin²56°+ 2tan18°tan72°-cot²30


=sin²34°+sin²(90°-34°)+2tan18°tan(90°-18°)-cot²30°
={sin²34°+cos²34°}+2{tan18°cot18°}-(√3)²
=1+2{1}-3
=3-3
=0

Answer 2

Answer:

0

Step-by-step explanation:

Given:- sin²34° + sin²56° + 2 tan18° tan72° - cot²30°

To Find:- Value of the given expression.

Solution:-

Given, sin²34° + sin²56° + 2 tan18° tan72° - cot²30°

= sin²34° + sin²(90 - 34)° + 2 tan18° tan(90 - 18)° - cot²30°  

= sin²34° + cos²34° + 2 tan18° cot18° - cot²30°   [∵ sin(90° - A) = cos A,

                                                                                      tan(90° - A) = cot A]

= 1 + 2 tan18° × $\frac{1}{tan18^{o} }$ - $(\sqrt{3} )^{2}$     [∵ sin²A + cos²A = 1 & cot30° = $\sqrt{3}$]

= 1 + 2 - 3

= 0

Therefore, sin²34° + sin²56° + 2 tan18° tan72° - cot²30° = 0.

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