Question

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Answer 1

Answer:

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Step-by-step explanation:

If 3 cot A = 4,  

cot A = 4 / 3

tan A = 3 / 4  

LHS

= 1 - tan^2 A / 1 + tan^2 A

= 1 - 3 / 4^2 / 1 + 3 / 4^2

= 1 - 9 / 16 / 1 + 9 / 16

= 16 - 9 / 16 + 9

= 7 / 25  

If tan A= 3 / 4

then sin A=3 / 5 (from trigonometric functions)

cos A= 4 / 5

So,apply these values in RHS

cos^2 A  -  sin^2 A

= 4 / 5^2  - 3 / 5^2

= 16 / 25  -  9 / 25

= 16 - 9 / 25

= 7 / 25

Hence RHS= LHS  

yes,for the given value both LHS =RHS

Answer 2

SOLUTION :-

=>Let ΔABC be a right triangle.

• 3cotA = 4

=> Hence, cotA = 4/3

=>We know that,

=> CotA= Side adjacent to ∠A / Side opposite of ∠A

=> CotA= AB/BC = 4/3

=>Hence, AB = 4 , BC=3 and AC = ?

=>By applying the Pythagoras theorem in ΔABC.

=> (AC)² = (AB)²+(BC)²

=> AC² = 4² + 3²

=>AC² = 16 + 9

=> AC² = 25

=> √(AC)² = √25 ---(Take square root on both sides.)

=> .°. AC = 5.

=> cosA = Side adjacent to ∠A / Hypotenuse

=> cosA = AB/AC

=> .°. cosA = 4/5

=>sinA = Side opposite of ∠A / Hypotenuse

=>sinA = BC / AC

=>.°. sinA = 3/5

=>tanA = Side opposite to ∠A / Side adjacent to ∠A

=>tanA = BC / AB

=> .°. tanA = 3/4

=> By substituting the above values of trigonometric functions in the LHS of the equation.

=>1-tan²A/1+tan²A = 1-(¾)²/1+(¾)²

=>1-tan²A/1+tan²A = 1 - 9/16 /1 + 9/16

=> =>1-tan²A/1+tan²A = 7/16 / 25/16

=>.°. 1-tan²A/1+tan²A = 7/25.

=> By substituting the above values of trigonometric functions in the RHS of the equation.

=>cos²A - sin²A = (4/5)² - (3/5)²

=>cos²A - sin²A = 16/25 - 9/25

=>.°. cos²A - sin²A = 7/25

=> .°. 1-tan²A/1+tan²A =cos²A - sin²A

=> Hence, it is proved.