7. Prove that tan B� sin B� + cos �B = sec B�.
11. Prove that tanλ cos^2λ +sin^2λ/sinλ = cos λ� + sin λ�.
12. Prove that 1+tanθ/1+tanθ = sec^2θ+2tanθ/ 1-tan^2θ.
21. Prove that sin^2w-cos^2w/ tan w sin w + cos w tan w = cos w� – cot w� cos w�.
tan*sin+cos
= sin^2/cos + cos
= (sin^2+cos^2)/cos
= 1/cos
= sec
(tan*cos^2 + sin^2)/sin
= (sin*cos + sin^2)/sin
= sin(cos+sin)/sin
= cos+sin
I think you mean
(1+tan)/(1-tan)
= (1+tan)^2/(1-tan^2)
= (1+2tan+tan^2)/(1-tan^2)
= (sec^2+2tan)/(1-tan^2)
the last one needs some parentheses. Too complicated and ambiguous
To prove these trigonometric identities, we will break down each identity into smaller steps and use basic trigonometric properties and identities to simplify and rearrange the expressions.
7. Prove that tan(B) sin(B) + cos(B) = sec(B):
Step 1: Start with the left-hand side (LHS) of the equation.
LHS = tan(B) sin(B) + cos(B)
Step 2: Expand the tangent function using sine and cosine.
LHS = (sin(B) / cos(B)) * sin(B) + cos(B)
Step 3: Simplify by multiplying and dividing by cos(B).
LHS = sin(B)^2 / cos(B) + cos(B)
Step 4: Convert the expression to a single fraction by finding a common denominator.
LHS = (sin(B)^2 + cos(B)^2) / cos(B)
Step 5: Use the Pythagorean identity sin^2(B) + cos^2(B) = 1.
LHS = 1 / cos(B)
Step 6: Replace cos(B) with its reciprocal, sec(B).
LHS = sec(B)
Thus, the left-hand side is equal to the right-hand side, and the identity is proven.
11. Prove that tan(λ) cos^2(λ) + sin^2(λ)/sin(λ) = cos(λ) + sin(λ):
Step 1: Start with the left-hand side (LHS) of the equation.
LHS = tan(λ) cos^2(λ) + (sin^2(λ))/sin(λ)
Step 2: Simplify the expression by factoring out common terms.
LHS = cos^2(λ) * (tan(λ) + sin(λ)/sin(λ))
Step 3: Simplify the expression inside the parentheses.
LHS = cos^2(λ) * (tan(λ) + 1)
Step 4: Combine like terms.
LHS = cos^2(λ) * tan(λ) + cos^2(λ)
Step 5: Use the Pythagorean identity sin^2(λ) + cos^2(λ) = 1 and rearrange the terms.
LHS = cos^2(λ) * tan(λ) + (1 - sin^2(λ))
Step 6: Simplify by distributing and combining like terms.
LHS = cos^2(λ) * tan(λ) + 1 - sin^2(λ)
Step 7: Use the Pythagorean identity 1 - sin^2(λ) = cos^2(λ).
LHS = cos^2(λ) * tan(λ) + cos^2(λ)
Step 8: Combine like terms.
LHS = cos^2(λ) * (tan(λ) + 1)
Step 9: Simplify the expression inside the parentheses.
LHS = cos^2(λ) * cos(λ)
Step 10: Use the identity cos^3(λ) = cos(λ).
LHS = cos(λ)
Thus, the left-hand side is equal to the right-hand side, and the identity is proven.
Note: The provided equation in question 12 seems to be incorrect. Could you please double-check the equation and provide the correct one?