Given that: 3 cot A = 4 ⇒ cot A = 4/3 Let cot A = 4k/3k, where k is a real number. In △ABC, by Pythagoras theorem, we have AC² = BC²+AB² = (3k)²+(4k)² = 9k²+16k² = 25k² ⇒ AC = √(25k²) = 5k Therefore, (1-tan²A)/(1+tan²A) = (1-(3/4)²)/(1+(3/4)²) = (1-(9/16))/(1+(9/16)) = ((16-9)/16)/((16+9)/16) = 7/25Read more
Given that: 3 cot A = 4
⇒ cot A = 4/3
Let cot A = 4k/3k, where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (3k)²+(4k)²
= 9k²+16k²
= 25k²
⇒ AC = √(25k²) = 5k
Therefore,
(1-tan²A)/(1+tan²A) = (1-(3/4)²)/(1+(3/4)²) = (1-(9/16))/(1+(9/16)) = ((16-9)/16)/((16+9)/16) = 7/25
and
cos² A-sin²A = (4/5)²-(3/5)² = 16/25 – 9/25 = (16-9)/25 = 7/25
Hence, (1-tan²A)/(1+tan²A) = cos²A-sin²A
Given that: tan A = 1/√3 Let tan A = 1k/√3, where k is a real number. In △ABC, by Pythagoras theorem, we have AC² = BC²+AB² = (1k)²+(V3k)² = k²+3k² = 4k² ⇒ AC = √(4k)² = 2k (i) sin A cos C+cos A sin C = 1/2 × 1/2 + √3/2 × √3/2 = 1/4 + 3/4 = (1+3)/4 = 4/4 = 1 (ii) cos A cos C sin A sin C = √3/2 × 1/2Read more
Given that: tan A = 1/√3
Let tan A = 1k/√3, where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (1k)²+(V3k)²
= k²+3k² = 4k²
⇒ AC = √(4k)² = 2k
(i) sin A cos C+cos A sin C
= 1/2 × 1/2 + √3/2 × √3/2 = 1/4 + 3/4 = (1+3)/4 = 4/4 = 1
(ii) cos A cos C sin A sin C
= √3/2 × 1/2 – 1/2 × √3/2 = √3/4 – √3/4 = 0
For better understanding of above question you can see here😃👇
Given that: in APQR, angle Q is right angled. Let QR x, therefore, PR = 25-x In △PQR, by Pythagoras theorem, we have PR² = PQ²+OQ² ⇒ (25-x)² = (5)²+(x)² ⇒ 625+x²-50x = 25+x² ⇒ 625-50x = 25 ⇒ 50x = 600 ⇒ x = 12 ⇒ QR = 12 Therefore, PR = 25-12 = 13 Now, sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13 andRead more
Given that: in APQR, angle Q is right angled.
Let QR x, therefore, PR = 25-x
In △PQR, by Pythagoras theorem, we have
PR² = PQ²+OQ²
⇒ (25-x)² = (5)²+(x)²
⇒ 625+x²-50x = 25+x²
⇒ 625-50x = 25
⇒ 50x = 600
⇒ x = 12
⇒ QR = 12
Therefore, PR = 25-12 = 13
Now, sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13 and tan P = QR/PQ = 12/5
If 3 cot A = 4, check whether 1- tan² A / 1+ tan² A= cos²A – sin² A or not.
Given that: 3 cot A = 4 ⇒ cot A = 4/3 Let cot A = 4k/3k, where k is a real number. In △ABC, by Pythagoras theorem, we have AC² = BC²+AB² = (3k)²+(4k)² = 9k²+16k² = 25k² ⇒ AC = √(25k²) = 5k Therefore, (1-tan²A)/(1+tan²A) = (1-(3/4)²)/(1+(3/4)²) = (1-(9/16))/(1+(9/16)) = ((16-9)/16)/((16+9)/16) = 7/25Read more
Given that: 3 cot A = 4
⇒ cot A = 4/3
Let cot A = 4k/3k, where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (3k)²+(4k)²
= 9k²+16k²
= 25k²
⇒ AC = √(25k²) = 5k
Therefore,
(1-tan²A)/(1+tan²A) = (1-(3/4)²)/(1+(3/4)²) = (1-(9/16))/(1+(9/16)) = ((16-9)/16)/((16+9)/16) = 7/25
and
cos² A-sin²A = (4/5)²-(3/5)² = 16/25 – 9/25 = (16-9)/25 = 7/25
Hence, (1-tan²A)/(1+tan²A) = cos²A-sin²A
Video Solution of the above question is here🙌😃
See lessIn triangle ABC, right-angled at B, if tan A = 1/√3 find the value of sin A cos C + cos A sin C.
Given that: tan A = 1/√3 Let tan A = 1k/√3, where k is a real number. In △ABC, by Pythagoras theorem, we have AC² = BC²+AB² = (1k)²+(V3k)² = k²+3k² = 4k² ⇒ AC = √(4k)² = 2k (i) sin A cos C+cos A sin C = 1/2 × 1/2 + √3/2 × √3/2 = 1/4 + 3/4 = (1+3)/4 = 4/4 = 1 (ii) cos A cos C sin A sin C = √3/2 × 1/2Read more
Given that: tan A = 1/√3
Let tan A = 1k/√3, where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (1k)²+(V3k)²
= k²+3k² = 4k²
⇒ AC = √(4k)² = 2k
(i) sin A cos C+cos A sin C
= 1/2 × 1/2 + √3/2 × √3/2 = 1/4 + 3/4 = (1+3)/4 = 4/4 = 1
(ii) cos A cos C sin A sin C
= √3/2 × 1/2 – 1/2 × √3/2 = √3/4 – √3/4 = 0
For better understanding of above question you can see here😃👇
See lessIn angle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Given that: in APQR, angle Q is right angled. Let QR x, therefore, PR = 25-x In △PQR, by Pythagoras theorem, we have PR² = PQ²+OQ² ⇒ (25-x)² = (5)²+(x)² ⇒ 625+x²-50x = 25+x² ⇒ 625-50x = 25 ⇒ 50x = 600 ⇒ x = 12 ⇒ QR = 12 Therefore, PR = 25-12 = 13 Now, sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13 andRead more
Given that: in APQR, angle Q is right angled.
Let QR x, therefore, PR = 25-x
In △PQR, by Pythagoras theorem, we have
PR² = PQ²+OQ²
⇒ (25-x)² = (5)²+(x)²
⇒ 625+x²-50x = 25+x²
⇒ 625-50x = 25
⇒ 50x = 600
⇒ x = 12
⇒ QR = 12
Therefore, PR = 25-12 = 13
Now, sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13 and tan P = QR/PQ = 12/5
Here is the video explanation of the question😺👇
See less