If a 2 2 - - b 2 tan 2 - - c 2 then prove that sin 2 - - c 2 - a 2 - c 2 - b 2-

a ^ 2 sec ^ 2 θ b ^ 2 tan ^ 2 θ = c ^ 2 a^2cos^2 θ b^2sin ^2 θcos^2 θ=c^2 a^2 b^2sin ^2 θ=c^2 cos^2 θ a^2 b^2sin ^2 θ=c^2(1 sin^2 θ) ...

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Trigonometric Identity- 1

If a 2 2 ...

Question

If $a^{2} {sec}^{2} θ - b^{2} {tan}^{2} θ = c^{2}$ then prove that ${sin}^{2} θ = \frac{c^{2} - a^{2}}{c^{2} - b^{2}}$ .

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\frac{a^{2} + b^{2} - c^{2}}{c^{2} + a^{2} - b^{2}} = \frac{tan B}{tan C}

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

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