Find the value of tan225°cot405° + tan765°cot675°:
Answer
669.3k+ views
Hint: These angles look absurd to you and you must be thinking, this is nowhere close to the 30°, 45°, 60° but here is a thing, if you look carefully to the angles you will find they are 2π + 45°, 4π ± 45° and$\dfrac{3\pi }{2}-{{45}^{0}}$. This way you can easily solve the expression.
Complete step by step answer:
We can write tan225° as tan (270° - 45°), cot405° as cot (360° + 45°), tan765° as tan (720° + 45°) and cot675° as cot (720° - 45°). Now, rewriting the expression given in question in terms of new angles as:
tan (270° - 45°)cot (360° + 45°) + tan (720° + 45°)cot(720° – 45°)
Now, 270° is$\dfrac{3\pi }{2}$, 360° is 2π and 720° is 4π. So we are replacing these degree angles in the form of multiples of π as follows:
$\tan \left( \dfrac{3\pi }{2}-{{45}^{0}} \right)\cot \left( 2\pi +{{45}^{0}} \right)+\tan \left( 4\pi +{{45}^{0}} \right)\cot \left( 4\pi -{{45}^{0}} \right)$
Now, we know that tan θ and cot θ are positive in the first and third quadrants and in the other quadrants, they are negative. So, opening the brackets will give expression as follows:
cot45°cot45° + tan45° (-cot45°)
From the trigonometric angles value, we know that both tan45° and cot45° have the value of 1. So, substituting these values in above expression we get:
1(1) + 1(-1)
$\Rightarrow $ 1 – 1
$\Rightarrow $ 0
Hence, the value of the given expression in the question is 0.
Note: Instead of writing the angles that I have written above you can also write angles as shown below:
225° = 180° + 45°
405° = 450° – 45°
765° = 810° – 45°
675° = 630° + 45°
This way of writing angles is also correct because these forms of angles are in the same quadrant as I have shown above so the tan θ and cot θ conversion won’t change.
Complete step by step answer:
We can write tan225° as tan (270° - 45°), cot405° as cot (360° + 45°), tan765° as tan (720° + 45°) and cot675° as cot (720° - 45°). Now, rewriting the expression given in question in terms of new angles as:
tan (270° - 45°)cot (360° + 45°) + tan (720° + 45°)cot(720° – 45°)
Now, 270° is$\dfrac{3\pi }{2}$, 360° is 2π and 720° is 4π. So we are replacing these degree angles in the form of multiples of π as follows:
$\tan \left( \dfrac{3\pi }{2}-{{45}^{0}} \right)\cot \left( 2\pi +{{45}^{0}} \right)+\tan \left( 4\pi +{{45}^{0}} \right)\cot \left( 4\pi -{{45}^{0}} \right)$
Now, we know that tan θ and cot θ are positive in the first and third quadrants and in the other quadrants, they are negative. So, opening the brackets will give expression as follows:
cot45°cot45° + tan45° (-cot45°)
From the trigonometric angles value, we know that both tan45° and cot45° have the value of 1. So, substituting these values in above expression we get:
1(1) + 1(-1)
$\Rightarrow $ 1 – 1
$\Rightarrow $ 0
Hence, the value of the given expression in the question is 0.
Note: Instead of writing the angles that I have written above you can also write angles as shown below:
225° = 180° + 45°
405° = 450° – 45°
765° = 810° – 45°
675° = 630° + 45°
This way of writing angles is also correct because these forms of angles are in the same quadrant as I have shown above so the tan θ and cot θ conversion won’t change.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

